OP Malhotra Class 12 Maths Solutions Chapter 14 Indefinite Integral 2 Exercise 14 (C)

S Chand Class 12 ICSE Maths Solutions Chapter 14 Indefinite Integral-2 Ex 14(c)

Question 1. \( \int x \sin 2 x d x \)
Answer: To solve this integral, we use the integration by parts method. We choose \(u = x\) and \(dv = \sin 2x dx\). Then we find \(du = dx\) and \(v = \int \sin 2x dx = -\frac{\cos 2x}{2}\). Applying the formula \( \int u dv = uv - \int v du \), we get \( -\frac{x \cos 2x}{2} - \int \left(-\frac{\cos 2x}{2}\right) dx \). We then integrate \( -\frac{\cos 2x}{2} \) to get \( \frac{\sin 2x}{4} \). So the final answer is \( -\frac{x \cos 2x}{2} + \frac{\sin 2x}{4} + C \). This technique helps us integrate products of functions.
In simple words: Whenever we have an integral with two different types of functions multiplied, like \( x \) and \( \sin 2x \), we use a special rule called integration by parts. After doing the steps, the answer is \( -\frac{x \cos 2x}{2} + \frac{\sin 2x}{4} + C \).

🎯 Exam Tip: Remember the integration by parts formula: \( \int u dv = uv - \int v du \). Correctly identifying \( u \) and \( dv \) is key to solving these problems.

Question 2. \( \int x \sec^2 x d x \)
Answer: We solve this integral using the integration by parts method. Let \(u = x\) and \(dv = \sec^2 x dx\). This gives us \(du = dx\) and \(v = \tan x\). When we put these into the integration by parts formula, we get \( x \tan x - \int \tan x dx \). The integral of \( \tan x \) is \( -\log |\cos x| \). So, the final result is \( x \tan x + \log |\cos x| + C \). Integration by parts is especially useful for products of algebraic and trigonometric functions.
In simple words: We use a rule to break down the integral. We choose one part to differentiate and another to integrate. After doing the math, the answer is \( x \tan x + \log |\cos x| + C \).

🎯 Exam Tip: For integration by parts, it's often helpful to remember the ILATE rule to choose \( u \): Inverse, Logarithmic, Algebraic, Trigonometric, Exponential. This helps in simplifying the integral.

Question 3. \( \int x^2 e^x d x \)
Answer: To evaluate this integral, we apply integration by parts two times. First, we choose \( u = x^2 \) and \( dv = e^x dx \), which leads to \( x^2 e^x - 2 \int x e^x dx \). Then, we apply integration by parts again for the term \( \int x e^x dx \), where \( u = x \) and \( dv = e^x dx \), giving us \( x e^x - e^x \). Substituting this back, the expression simplifies to \( x^2 e^x - 2(x e^x - e^x) + C \). Finally, this gives us \( (x^2 - 2x + 2) e^x + C \). This nested application of the rule is common for powers of x multiplied by exponentials.
In simple words: We solve this integral by using the integration by parts rule two times. We handle the \( x^2 \) part first, then the \( x \) part. When we combine everything, the final answer is \( (x^2 - 2x + 2) e^x + C \).

🎯 Exam Tip: When you have \( x^n e^x \) or \( x^n \sin x \) type integrals, you'll generally apply integration by parts \( n \) times. Be careful with signs in each step.

Question 4. \( \int x \cos 2x d x \)
Answer: We evaluate this integral using the integration by parts method. We select \(u = x\) and \(dv = \cos 2x dx\). This means \(du = dx\) and \(v = \frac{\sin 2x}{2}\). Applying the formula, we get \( \frac{x \sin 2x}{2} - \int \frac{\sin 2x}{2} dx \). Next, we integrate \( \frac{\sin 2x}{2} \) to get \( -\frac{\cos 2x}{4} \). After combining terms, the final answer is \( \frac{x \sin 2x}{2} + \frac{\cos 2x}{4} + C \). Understanding basic trigonometric integrals is crucial for these problems.
In simple words: We use the integration by parts rule. We integrate one part and differentiate the other. The calculation gives us \( \frac{x \sin 2x}{2} + \frac{\cos 2x}{4} + C \) as the final answer.

🎯 Exam Tip: Always pay attention to the constant of integration \( +C \) at the end of every indefinite integral. It represents the family of all possible antiderivatives.

Question 5. \( \int (1+x) \cos 2x d x \)
Answer: We solve this integral using the integration by parts formula. We assign \( u = (1+x) \) and \( dv = \cos 2x dx \). From this, we find \( du = dx \) and \( v = \frac{\sin 2x}{2} \). Applying the integration by parts formula gives us \( (1+x) \frac{\sin 2x}{2} - \int \frac{\sin 2x}{2} dx \). The remaining integral \( \int \sin 2x dx \) becomes \( -\frac{\cos 2x}{2} \). Combining these parts, the final answer is \( (1+x) \frac{\sin 2x}{2} + \frac{\cos 2x}{4} + C \). This shows how the product rule for differentiation is reversed to find integrals.
In simple words: We use the integration by parts rule for this problem. We break it into two parts and integrate each carefully. After performing the steps, the answer is \( (1+x) \frac{\sin 2x}{2} + \frac{\cos 2x}{4} + C \).

🎯 Exam Tip: When applying integration by parts with a polynomial term, the polynomial usually becomes \( u \) because its derivative simplifies in subsequent steps.

Question 6. \( \int (1-x^2) \log x d x \)
Answer: To solve this integral, we use the integration by parts rule. We choose \( u = \log x \) because it simplifies when differentiated, and \( dv = (1-x^2) dx \). Integrating \( dv \) gives \( v = x - \frac{x^3}{3} \). Applying the formula, we get \( \left(x - \frac{x^3}{3}\right) \log x - \int \left(x - \frac{x^3}{3}\right) \frac{1}{x} dx \). We then simplify the integral term to \( \int \left(1 - \frac{x^2}{3}\right) dx \), which integrates to \( x - \frac{x^3}{9} \). Thus, the final answer is \( \left(x - \frac{x^3}{3}\right) \log x - x + \frac{x^3}{9} + C \). This demonstrates the power of choosing the right parts for integration.
In simple words: We use the integration by parts technique for this problem. We differentiate the log part and integrate the polynomial part. After simplifying, the final answer is \( \left(x - \frac{x^3}{3}\right) \log x - x + \frac{x^3}{9} + C \).

🎯 Exam Tip: When integrating a product involving a logarithmic function, it's almost always best to choose the logarithmic function as \( u \) because its derivative is algebraic and simpler.

Question 7. \( \int x^2 \sin x d x \)
Answer: To solve this integral, we apply integration by parts two times. First, we set \( u = x^2 \) and \( dv = \sin x dx \), which gives us \( -x^2 \cos x + 2 \int x \cos x dx \). For the remaining integral \( \int x \cos x dx \), we apply integration by parts again, choosing \( u = x \) and \( dv = \cos x dx \). This second step yields \( x \sin x + \cos x \). Substituting this back into the main equation gives \( -x^2 \cos x + 2(x \sin x + \cos x) + C \). The final simplified answer is \( -x^2 \cos x + 2x \sin x + 2 \cos x + C \). This method is a robust way to integrate products involving powers of \( x \) and trigonometric functions.
In simple words: We solve this integral by using the integration by parts rule two times. We handle the \( x^2 \) part first, then the \( x \) part. After all the steps, the answer is \( -x^2 \cos x + 2x \sin x + 2 \cos x + C \).

🎯 Exam Tip: Always be careful with the signs when dealing with derivatives and integrals of trigonometric functions, especially \( \sin x \) and \( \cos x \).

Question 8. \( \int (\log x)^2 d x \)
Answer: To solve this integral, we apply integration by parts two times. First, we set \( u = (\log x)^2 \) and \( dv = 1 dx \), which simplifies to \( x (\log x)^2 - 2 \int \log x dx \). For the remaining integral \( \int \log x dx \), we apply integration by parts again, choosing \( u = \log x \) and \( dv = 1 dx \). This yields \( x \log x - x \). Substituting this back into the main equation gives \( x (\log x)^2 - 2 (x \log x - x) + C \). The final simplified answer is \( x (\log x)^2 - 2x \log x + 2x + C \). This method is useful for integrating powers of logarithmic functions.
In simple words: We use the integration by parts rule two times for this problem. We first handle the \( (\log x)^2 \) part and then the \( \log x \) part. After all the steps, the answer is \( x (\log x)^2 - 2x \log x + 2x + C \).

🎯 Exam Tip: Remember the standard integral \( \int \log x dx = x \log x - x + C \). This result is frequently used as a step in more complex integration problems.

Question 9. \( \int 2x e^{5x} d x \)
Answer: To solve this integral, we apply integration by parts. We choose \( u = 2x \) and \( dv = e^{5x} dx \). From this, we find \( du = 2 dx \) and \( v = \frac{e^{5x}}{5} \). Applying the integration by parts formula, we get \( 2x \left(\frac{e^{5x}}{5}\right) - \int \left(\frac{e^{5x}}{5}\right) 2 dx \). Simplifying, this becomes \( \frac{2x e^{5x}}{5} - \frac{2}{5} \int e^{5x} dx \). The integral of \( e^{5x} \) is \( \frac{e^{5x}}{5} \). So, after combining terms and factoring, the final answer is \( \frac{2e^{5x}}{25} (5x - 1) + C \). This method is very effective for products of algebraic and exponential functions.
In simple words: We use the integration by parts rule here. We differentiate the \( 2x \) part and integrate the \( e^{5x} \) part. After calculation, the final answer is \( \frac{2e^{5x}}{25} (5x - 1) + C \).

🎯 Exam Tip: When integrating \( e^{ax} \), remember that \( \int e^{ax} dx = \frac{e^{ax}}{a} + C \). This is a common point where students might make a mistake.

Question 10. \( \int \frac{\log \left(1+x^2\right)}{x^2} d x \)
Answer: To solve this integral, we use the integration by parts rule. We choose \( u = \log(1+x^2) \) and \( dv = x^{-2} dx \). From this, we find \( du = \frac{2x}{1+x^2} dx \) and \( v = -\frac{1}{x} \). Applying the formula, we get \( -\frac{\log(1+x^2)}{x} - \int \left(-\frac{1}{x}\right) \left(\frac{2x}{1+x^2}\right) dx \). This simplifies to \( -\frac{\log(1+x^2)}{x} + 2 \int \frac{1}{1+x^2} dx \). The integral of \( \frac{1}{1+x^2} \) is \( \tan^{-1} x \). So, the final answer is \( -\frac{\log(1+x^2)}{x} + 2 \tan^{-1} x + C \). This problem combines logarithmic and inverse trigonometric function properties.
In simple words: We solve this integral using the integration by parts rule. We differentiate the log part and integrate the \( x^{-2} \) part. After all the steps, the final answer is \( -\frac{\log(1+x^2)}{x} + 2 \tan^{-1} x + C \).

🎯 Exam Tip: Remember the standard integral \( \int \frac{1}{1+x^2} dx = \tan^{-1} x + C \). This is a very common result in integration problems.

Question 11. \( \int x^2 \log x d x \)
Answer: To solve this integral, we use the integration by parts method. We choose \( u = \log x \) because its derivative simplifies, and \( dv = x^2 dx \). This gives us \( du = \frac{1}{x} dx \) and \( v = \frac{x^3}{3} \). Applying the formula, we get \( \frac{x^3}{3} \log x - \int \left(\frac{x^3}{3}\right) \frac{1}{x} dx \). This simplifies to \( \frac{x^3}{3} \log x - \frac{1}{3} \int x^2 dx \). After integrating \( x^2 \), the term becomes \( \frac{x^3}{9} \). So, the final answer is \( \frac{x^3}{3} \log x - \frac{x^3}{9} + C \). This is a classic example of integrating a logarithmic function with a polynomial.
In simple words: We use the integration by parts rule. We choose the log part to differentiate and the \( x^2 \) part to integrate. After performing the steps, the final answer is \( \frac{x^3}{3} \log x - \frac{x^3}{9} + C \).

🎯 Exam Tip: When dealing with integrals of the form \( \int x^n \log x dx \), it's standard practice to let \( u = \log x \) and \( dv = x^n dx \) for effective integration by parts.

Question 12. \( \int x^3 \log_e x^2 d x \)
Answer: First, we simplify the logarithm using the property \( \log x^2 = 2 \log x \). So, the integral becomes \( 2 \int x^3 \log x dx \). Next, we apply integration by parts to \( \int x^3 \log x dx \). We set \( u = \log x \) and \( dv = x^3 dx \). This leads to \( du = \frac{1}{x} dx \) and \( v = \frac{x^4}{4} \). Applying the formula, we get \( \frac{x^4}{4} \log x - \int \frac{x^3}{4} dx \). Integrating \( \frac{x^3}{4} \) gives \( \frac{x^4}{16} \). Multiplying the entire expression by 2 (from the initial simplification), the final answer is \( \frac{x^4}{2} \log x - \frac{x^4}{8} + C \). This problem highlights simplifying the integrand before applying integration techniques.
In simple words: We first simplify the logarithm in the question. Then we use the integration by parts rule. We differentiate the log part and integrate the \( x^3 \) part. After all the steps, the final answer is \( \frac{x^4}{2} \log x - \frac{x^4}{8} + C \).

🎯 Exam Tip: Always simplify logarithmic expressions in the integrand using properties like \( \log a^b = b \log a \) before applying integration techniques, as it often makes the problem much easier.

Question 13. \( \int x \sin^2 3x d x \)
Answer: First, we use the trigonometric identity \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \) to rewrite \( \sin^2 3x \) as \( \frac{1 - \cos 6x}{2} \). This transforms the integral into \( \frac{1}{2} \int (x - x \cos 6x) dx \). We integrate \( x \) directly to get \( \frac{x^2}{2} \). For \( \int x \cos 6x dx \), we apply integration by parts. Letting \( u = x \) and \( dv = \cos 6x dx \), we find this integral is \( \frac{x \sin 6x}{6} + \frac{\cos 6x}{36} \). Combining all parts and multiplying by \( \frac{1}{2} \), the final answer is \( \frac{x^2}{4} - \frac{x \sin 6x}{12} - \frac{\cos 6x}{72} + C \). This illustrates how trigonometric identities can simplify integrals before applying integration by parts.
In simple words: We first change \( \sin^2 3x \) using a math rule. Then, we use integration by parts for one part of the new integral. After doing all the steps, the final answer is \( \frac{x^2}{4} - \frac{x \sin 6x}{12} - \frac{\cos 6x}{72} + C \).

🎯 Exam Tip: Always look for opportunities to simplify trigonometric functions in the integrand using identities before applying integration techniques. This often reduces the complexity of the problem significantly.

Question 14. \( \int \frac{x - \sin x}{1 - \cos x} d x \)
Answer: First, we use trigonometric half-angle identities to simplify the integrand. We replace \( \sin x \) with \( 2 \sin \frac{x}{2} \cos \frac{x}{2} \) and \( 1 - \cos x \) with \( 2 \sin^2 \frac{x}{2} \). The integral then becomes \( \int \left( \frac{x}{2} \operatorname{cosec}^2 \frac{x}{2} - \cot \frac{x}{2} \right) dx \). We separate this into two integrals: \( \frac{1}{2} \int x \operatorname{cosec}^2 \frac{x}{2} dx \) and \( - \int \cot \frac{x}{2} dx \). For the first integral, we apply integration by parts, letting \( u = x \) and \( dv = \operatorname{cosec}^2 \frac{x}{2} dx \). This part results in \( -2x \cot \frac{x}{2} + 4 \log |\sin \frac{x}{2}| \). The second integral is \( -2 \log |\sin \frac{x}{2}| \). Combining all parts and simplifying, the \( \log |\sin \frac{x}{2}| \) terms cancel out. The final answer is \( -x \cot \frac{x}{2} + C \). This problem showcases simplifying complex fractions with trigonometric identities.
In simple words: We first use special math rules for sine and cosine to make the integral simpler. Then, we use the integration by parts rule for one part. After putting everything together, the final answer is \( -x \cot \frac{x}{2} + C \).

🎯 Exam Tip: When you see expressions like \( 1 \pm \cos x \) or \( \sin x \) in fractions, always consider using half-angle trigonometric identities to simplify the integrand before attempting integration.

Question 15. \( \int \cos^{-1} x d x \)
Answer: To integrate \( \cos^{-1} x \), we use integration by parts, treating it as \( \int \cos^{-1} x \cdot 1 dx \). We set \( u = \cos^{-1} x \) and \( dv = 1 dx \). This gives us \( du = -\frac{1}{\sqrt{1-x^2}} dx \) and \( v = x \). Applying the integration by parts formula yields \( x \cos^{-1} x - \int x \left(-\frac{1}{\sqrt{1-x^2}}\right) dx \), which simplifies to \( x \cos^{-1} x + \int \frac{x}{\sqrt{1-x^2}} dx \). For the remaining integral, we use a substitution \( t = 1-x^2 \), which simplifies the integral to \( -\sqrt{1-x^2} \). Combining these, the final answer is \( x \cos^{-1} x - \sqrt{1-x^2} + C \). This method is widely applicable for inverse trigonometric functions.
In simple words: We integrate \( \cos^{-1} x \) using the integration by parts rule, thinking of it as \( \cos^{-1} x \) times 1. After doing the steps and a small substitution, the final answer is \( x \cos^{-1} x - \sqrt{1-x^2} + C \).

🎯 Exam Tip: When integrating inverse trigonometric functions like \( \sin^{-1} x \), \( \cos^{-1} x \), \( \tan^{-1} x \), always use integration by parts by multiplying by 1 and setting the inverse function as \( u \).

Question 16. \( \int \cot^{-1} x d x \)
Answer: To integrate \( \cot^{-1} x \), we use integration by parts, considering it as \( \int \cot^{-1} x \cdot 1 dx \). We assign \( u = \cot^{-1} x \) and \( dv = 1 dx \). This yields \( du = -\frac{1}{1+x^2} dx \) and \( v = x \). Applying the integration by parts formula, we get \( x \cot^{-1} x - \int x \left(-\frac{1}{1+x^2}\right) dx \), which simplifies to \( x \cot^{-1} x + \int \frac{x}{1+x^2} dx \). For the remaining integral, we use a substitution \( t = 1+x^2 \), which gives \( \frac{1}{2} \log (1+x^2) \). Combining these parts, the final answer is \( x \cot^{-1} x + \frac{1}{2} \log (1+x^2) + C \). This demonstrates the integration of another inverse trigonometric function.
In simple words: We integrate \( \cot^{-1} x \) using the integration by parts rule, thinking of it as \( \cot^{-1} x \) times 1. After doing the steps and a small substitution, the final answer is \( x \cot^{-1} x + \frac{1}{2} \log (1+x^2) + C \).

🎯 Exam Tip: When integrating \( \frac{f'(x)}{f(x)} \), the result is \( \log |f(x)| + C \). This is a helpful pattern to recognize for integrals after applying integration by parts.

Question 17. \( \int \operatorname{cosec}^{-1} x d x \text{ if } x > 0 \)
Answer: To integrate \( \operatorname{cosec}^{-1} x \), we use integration by parts, considering it as \( \int \operatorname{cosec}^{-1} x \cdot 1 dx \). We set \( u = \operatorname{cosec}^{-1} x \) and \( dv = 1 dx \). Since \( x > 0 \), we find \( du = -\frac{1}{x \sqrt{x^2-1}} dx \) and \( v = x \). Applying the integration by parts formula, we get \( x \operatorname{cosec}^{-1} x - \int x \left(-\frac{1}{x \sqrt{x^2-1}}\right) dx \), which simplifies to \( x \operatorname{cosec}^{-1} x + \int \frac{1}{\sqrt{x^2-1}} dx \). The integral \( \int \frac{1}{\sqrt{x^2-1}} dx \) is a standard form that equals \( \log |x + \sqrt{x^2-1}| \). Combining these, the final answer is \( x \operatorname{cosec}^{-1} x + \log |x + \sqrt{x^2-1}| + C \). This is a standard method for integrating inverse hyperbolic functions.
In simple words: We integrate \( \operatorname{cosec}^{-1} x \) using the integration by parts rule. We treat it as \( \operatorname{cosec}^{-1} x \) times 1. After doing the steps, the final answer is \( x \operatorname{cosec}^{-1} x + \log |x + \sqrt{x^2-1}| + C \).

🎯 Exam Tip: Memorize common standard integral formulas, such as \( \int \frac{1}{\sqrt{x^2-a^2}} dx = \log |x + \sqrt{x^2-a^2}| + C \), as they frequently appear in integration problems.

Question 18. \( \int x^2 \cos^{-1} x d x \)
Answer: To solve this integral, we apply integration by parts. We set \( u = \cos^{-1} x \) and \( dv = x^2 dx \). This gives us \( du = -\frac{1}{\sqrt{1-x^2}} dx \) and \( v = \frac{x^3}{3} \). Applying the formula, we obtain \( \frac{x^3}{3} \cos^{-1} x + \frac{1}{3} \int \frac{x^3}{\sqrt{1-x^2}} dx \). For the integral \( \int \frac{x^3}{\sqrt{1-x^2}} dx \), we use the substitution \( t = 1-x^2 \). This transforms the integral into \( -\frac{1}{2} \int (t^{-1/2} - t^{1/2}) dt \), which evaluates to \( -\sqrt{1-x^2} + \frac{1}{3}(1-x^2)^{3/2} \). Substituting this back, the final answer is \( \frac{x^3}{3} \cos^{-1} x - \frac{1}{3}\sqrt{1-x^2} + \frac{1}{9}(1-x^2)^{3/2} + C \). This demonstrates a complex application of integration by parts and substitution.
In simple words: We solve this integral using the integration by parts rule. We differentiate the \( \cos^{-1} x \) part and integrate the \( x^2 \) part. Then, for the new integral, we use a substitution. After all the calculations, the final answer is \( \frac{x^3}{3} \cos^{-1} x - \frac{1}{3}\sqrt{1-x^2} + \frac{1}{9}(1-x^2)^{3/2} + C \).

🎯 Exam Tip: For integrals of the form \( \int x^n \sin^{-1} x dx \) or \( \int x^n \cos^{-1} x dx \), always choose the inverse trigonometric function as \( u \) and \( x^n dx \) as \( dv \). The resulting integral often requires a trigonometric or algebraic substitution.

Question 19. \( \int \sin \sqrt{x} d x \)
Answer: To solve this integral, we first use a substitution. Let \( y = \sqrt{x} \). This means \( x = y^2 \) and \( dx = 2y dy \). The integral transforms to \( 2 \int y \sin y dy \). Next, we apply integration by parts to this new integral. We set \( u = y \) and \( dv = \sin y dy \), which gives \( du = dy \) and \( v = -\cos y \). Applying the formula yields \( y(-\cos y) - \int (-\cos y) dy \), which simplifies to \( -y \cos y + \sin y \). Finally, substituting back \( y = \sqrt{x} \), the integral becomes \( 2 (-\sqrt{x} \cos \sqrt{x} + \sin \sqrt{x}) + C \). This illustrates using substitution to make an integral suitable for integration by parts.
In simple words: We first change \( \sqrt{x} \) to a new variable. This makes the integral easier to handle. Then we use the integration by parts rule. After putting everything back, the final answer is \( 2 (-\sqrt{x} \cos \sqrt{x} + \sin \sqrt{x}) + C \).

🎯 Exam Tip: If the argument of a trigonometric or exponential function is not linear (e.g., \( \sqrt{x} \), \( x^2 \)), a substitution is often the first step to simplify the integral before applying other techniques like integration by parts.

Question 20. \( \int \tan^{-1} \sqrt{x} d x \)
Answer: First, we perform a substitution. Let \( y = \sqrt{x} \), which implies \( x = y^2 \) and \( dx = 2y dy \). The integral becomes \( 2 \int y \tan^{-1} y dy \). Next, we use integration by parts for this new integral. We choose \( u = \tan^{-1} y \) and \( dv = y dy \), which gives \( du = \frac{1}{1+y^2} dy \) and \( v = \frac{y^2}{2} \). Applying the formula, we get \( \frac{y^2}{2} \tan^{-1} y - \frac{1}{2} \int \frac{y^2}{1+y^2} dy \). The integral \( \int \frac{y^2}{1+y^2} dy \) simplifies to \( y - \tan^{-1} y \). Substituting back, we obtain \( y^2 \tan^{-1} y - y + \tan^{-1} y \). Finally, substituting \( y = \sqrt{x} \), the solution is \( (x+1) \tan^{-1} \sqrt{x} - \sqrt{x} + C \). This demonstrates a common pattern for integrating inverse trigonometric functions with root arguments.
In simple words: We first use a substitution for \( \sqrt{x} \). Then, we use the integration by parts rule. We choose the \( \tan^{-1} \) part to differentiate and the \( y \) part to integrate. After simplifying and putting the original \( x \) back, the final answer is \( (x+1) \tan^{-1} \sqrt{x} - \sqrt{x} + C \).

🎯 Exam Tip: Integrals involving \( \tan^{-1} \sqrt{x} \) or similar forms often require an initial substitution \( t = \sqrt{x} \) followed by integration by parts. Remember that \( \int \frac{x^2}{1+x^2} dx = x - \tan^{-1} x + C \).

Question 21. \( \int \frac{\log x}{x^2} d x \)
Answer: To solve this integral, we apply integration by parts. We choose \( u = \log x \) because its derivative simplifies, and \( dv = x^{-2} dx \). This yields \( du = \frac{1}{x} dx \) and \( v = -\frac{1}{x} \). Applying the formula gives \( \log x \left(-\frac{1}{x}\right) - \int \left(-\frac{1}{x}\right) \left(\frac{1}{x}\right) dx \). This simplifies to \( -\frac{\log x}{x} + \int x^{-2} dx \). Integrating \( x^{-2} \) results in \( -\frac{1}{x} \). So, the final answer is \( -\frac{\log x}{x} - \frac{1}{x} + C \), which can also be written as \( -\frac{1}{x} (\log x + 1) + C \). This is a common form when integrating logarithmic functions with negative powers of x.
In simple words: We use the integration by parts rule. We differentiate the \( \log x \) part and integrate the \( x^{-2} \) part. After the calculations, the final answer is \( -\frac{1}{x} (\log x + 1) + C \).

🎯 Exam Tip: Always simplify the integrand before starting. Here, writing \( \frac{\log x}{x^2} \) as \( \log x \cdot x^{-2} \) makes it clear which parts to choose for integration by parts.

Question 22. \( \int x \sin^{-1} x d x \)
Answer: To solve this integral, we apply integration by parts. We choose \( u = \sin^{-1} x \) and \( dv = x dx \). This gives us \( du = \frac{1}{\sqrt{1-x^2}} dx \) and \( v = \frac{x^2}{2} \). Applying the formula, we obtain \( \frac{x^2}{2} \sin^{-1} x - \frac{1}{2} \int \frac{x^2}{\sqrt{1-x^2}} dx \). For the integral \( \int \frac{x^2}{\sqrt{1-x^2}} dx \), we use the trigonometric substitution \( x = \sin \theta \). This transforms the integral into \( \int \sin^2 \theta d\theta \), which simplifies to \( \frac{1}{2}(\theta - \sin \theta \cos \theta) \). Substituting back into terms of \( x \), this part becomes \( \frac{1}{2}(\sin^{-1} x - x \sqrt{1-x^2}) \). Combining all terms, the final answer is \( \frac{1}{4} (2x^2-1) \sin^{-1} x + \frac{x \sqrt{1-x^2}}{4} + C \). This demonstrates a complex integration involving inverse trigonometric functions.
In simple words: We use the integration by parts rule. We differentiate the \( \sin^{-1} x \) part and integrate the \( x \) part. Then, for the new integral, we use a trigonometric substitution to solve it. After putting everything back and simplifying, the final answer is \( \frac{1}{4} (2x^2-1) \sin^{-1} x + \frac{x \sqrt{1-x^2}}{4} + C \).

🎯 Exam Tip: Integrals of \( \frac{x^n}{\sqrt{a^2-x^2}} \) often require trigonometric substitution (like \( x = a \sin \theta \) or \( x = a \cos \theta \)) to simplify the integrand before integration.

Question 23. \( \int \frac{\log (\log x)}{x} d x \)
Answer: To solve this integral, we first use a substitution. Let \( y = \log x \). Then, \( dy = \frac{1}{x} dx \). The integral transforms to \( \int \log y dy \). Next, we apply integration by parts to this new integral, treating it as \( \int \log y \cdot 1 dy \). We set \( u = \log y \) and \( dv = 1 dy \), which gives \( du = \frac{1}{y} dy \) and \( v = y \). Applying the formula yields \( y \log y - \int y \left(\frac{1}{y}\right) dy \), which simplifies to \( y \log y - y \). Finally, substituting back \( y = \log x \), the integral becomes \( \log x (\log(\log x) - 1) + C \). This is a standard integral involving nested logarithmic functions.
In simple words: We first use a substitution for \( \log x \). This makes the integral simpler. Then we use the integration by parts rule. After putting everything back, the final answer is \( \log x (\log(\log x) - 1) + C \).

🎯 Exam Tip: For integrals of \( \log(\log x)/x \) or similar forms, the substitution \( t = \log x \) is almost always the correct first step to simplify the integral into a known form like \( \int \log t dt \).

Question 24. \( \int \sin^3 \sqrt{x} d x \)
Answer: To solve this integral, we first use the substitution \( y = \sqrt{x} \), so \( x = y^2 \) and \( dx = 2y dy \). This transforms the integral into \( 2 \int y \sin^3 y dy \). Next, we use the trigonometric identity \( \sin^3 y = \frac{3 \sin y - \sin 3y}{4} \) to rewrite the integrand. The integral becomes \( \frac{1}{2} \int (3y \sin y - y \sin 3y) dy \). We then split this into two integrals and apply integration by parts to each: \( \int y \sin y dy \) and \( \int y \sin 3y dy \). After integrating both parts and combining the results, and substituting back \( y = \sqrt{x} \), the final answer is \( -\frac{3}{2} \sqrt{x} \cos \sqrt{x} + \frac{3}{2} \sin \sqrt{x} + \frac{\sqrt{x} \cos 3\sqrt{x}}{6} - \frac{\sin 3\sqrt{x}}{18} + C \). This problem requires a combination of substitution, trigonometric identities, and integration by parts.
In simple words: We first use a substitution for \( \sqrt{x} \). Then, we use a trigonometric identity to change \( \sin^3 y \). After that, we use the integration by parts rule two times for the different parts of the integral. Finally, after putting \( \sqrt{x} \) back, the answer is \( -\frac{3}{2} \sqrt{x} \cos \sqrt{x} + \frac{3}{2} \sin \sqrt{x} + \frac{\sqrt{x} \cos 3\sqrt{x}}{6} - \frac{\sin 3\sqrt{x}}{18} + C \).

🎯 Exam Tip: For higher powers of sine or cosine in an integrand, always try to reduce the power using trigonometric identities (like triple angle or half-angle formulas) before applying other integration techniques. Be precise with coefficient and sign handling.

Question 25. \( \int \operatorname{cosec}^3 x d x \)
Answer: To integrate \( \operatorname{cosec}^3 x \), we use integration by parts, setting \( u = \operatorname{cosec} x \) and \( dv = \operatorname{cosec}^2 x dx \). This gives \( du = -\operatorname{cosec} x \cot x dx \) and \( v = -\cot x \). Applying the formula, we obtain \( -\operatorname{cosec} x \cot x - \int (-\cot x)(-\operatorname{cosec} x \cot x) dx \). This simplifies to \( -\operatorname{cosec} x \cot x - \int \operatorname{cosec} x \cot^2 x dx \). Using the identity \( \cot^2 x = \operatorname{cosec}^2 x - 1 \), we can split the integral, and observe that the original integral \( I \) reappears on the right side. This leads to \( 2I = -\operatorname{cosec} x \cot x + \int \operatorname{cosec} x dx \). We know \( \int \operatorname{cosec} x dx = \log |\tan \frac{x}{2}| \). Therefore, the final answer is \( I = -\frac{1}{2} \operatorname{cosec} x \cot x + \frac{1}{2} \log |\tan \frac{x}{2}| + C \). This is a classic example of a reduction formula integral.
In simple words: We integrate \( \operatorname{cosec}^3 x \) using the integration by parts rule. After the first step, we use a trigonometric identity. This makes the original integral appear again on the right side. We then solve for \( I \). The final answer is \( -\frac{1}{2} \operatorname{cosec} x \cot x + \frac{1}{2} \log |\tan \frac{x}{2}| + C \).

🎯 Exam Tip: For integrals of \( \sin^n x \), \( \cos^n x \), \( \sec^n x \), or \( \operatorname{cosec}^n x \) where \( n \) is odd and greater than 1, integration by parts combined with trigonometric identities often leads to a recursive formula (reduction formula) where the original integral reappears.

Question 26. \( \int x \cos^3 x d x \)
Answer: First, we use the trigonometric identity \( \cos^3 x = \frac{\cos 3x + 3 \cos x}{4} \) to simplify the integrand. The integral becomes \( \frac{1}{4} \int (x \cos 3x + 3x \cos x) dx \). We then split this into two separate integrals and apply integration by parts to each. For \( \int x \cos 3x dx \), we get \( \frac{x \sin 3x}{3} + \frac{\cos 3x}{9} \). For \( \int x \cos x dx \), we get \( x \sin x + \cos x \). Combining these results and distributing the \( \frac{1}{4} \), the final answer is \( \frac{x \sin 3x}{12} + \frac{\cos 3x}{36} + \frac{3x \sin x}{4} + \frac{3 \cos x}{4} + C \). This integral shows how useful trigonometric identities are for making products of functions integrable by parts.
In simple words: We first change \( \cos^3 x \) using a math rule. Then, we split the integral and use the integration by parts rule two times. After doing all the steps, the final answer is \( \frac{x \sin 3x}{12} + \frac{\cos 3x}{36} + \frac{3x \sin x}{4} + \frac{3 \cos x}{4} + C \).

🎯 Exam Tip: Always simplify powers of trigonometric functions using identities (like triple angle or double angle) to reduce them to linear terms before applying integration by parts, especially when multiplied by algebraic terms.

Question 27. \( \int x \tan^2 x d x \)
Answer: First, we use the trigonometric identity \( \tan^2 x = \sec^2 x - 1 \) to rewrite the integrand. This transforms the integral into \( \int x (\sec^2 x - 1) dx \), which splits into \( \int x \sec^2 x dx - \int x dx \). We know \( \int x dx = \frac{x^2}{2} \). For \( \int x \sec^2 x dx \), we apply integration by parts. We choose \( u = x \) and \( dv = \sec^2 x dx \), which gives \( x \tan x - \int \tan x dx \). The integral of \( \tan x \) is \( -\log |\cos x| \). Combining all these parts, the final answer is \( x \tan x + \log |\cos x| - \frac{x^2}{2} + C \). This problem shows how identities can simplify products for integration by parts.
In simple words: We first change \( \tan^2 x \) using a math rule. Then, we use the integration by parts rule for one part of the integral. After all the steps, the final answer is \( x \tan x + \log |\cos x| - \frac{x^2}{2} + C \).

🎯 Exam Tip: Whenever you encounter \( \tan^2 x \) or \( \cot^2 x \) in an integral, immediately use the identities \( \tan^2 x = \sec^2 x - 1 \) and \( \cot^2 x = \operatorname{cosec}^2 x - 1 \) to simplify the expression, as their derivatives are standard integrals.

Question 28. \( \int x\left(\frac{\sec 2 x-1}{\sec 2 x+1}\right) d x \)
Answer: First, we simplify the trigonometric fraction. We rewrite \( \sec 2x \) as \( \frac{1}{\cos 2x} \). The expression becomes \( \frac{1 - \cos 2x}{1 + \cos 2x} \). Using the half-angle identities \( 1 - \cos 2x = 2 \sin^2 x \) and \( 1 + \cos 2x = 2 \cos^2 x \), this simplifies to \( \frac{2 \sin^2 x}{2 \cos^2 x} = \tan^2 x \). So, the integral is transformed to \( \int x \tan^2 x d x \). This is the same integral as Question 27. Therefore, we use the identity \( \tan^2 x = \sec^2 x - 1 \) and apply integration by parts to \( \int x \sec^2 x dx \). The final answer is \( x \tan x - \log |\sec x| - \frac{x^2}{2} + C \). Recognizing trigonometric simplifications at the start is key.
In simple words: We first simplify the fraction using special trigonometry rules. It turns out to be \( \tan^2 x \). So the question becomes the same as Question 27. We then use the integration by parts rule. After all the steps, the final answer is \( x \tan x - \log |\sec x| - \frac{x^2}{2} + C \).

🎯 Exam Tip: Always simplify complex trigonometric fractions first, often using identities relating to \( 1 \pm \cos \theta \) or \( \sec \theta \), as they frequently simplify to powers of \( \tan \theta \) or \( \cot \theta \).

Question 29. \( \int \tan^{-1}\left(\frac{2 x}{1-x^2}\right) d x \)
Answer: First, we use the inverse trigonometric identity \( \tan^{-1} \left(\frac{2x}{1-x^2}\right) = 2 \tan^{-1} x \). This transforms the integral into \( 2 \int \tan^{-1} x d x \). Next, we apply integration by parts to \( \int \tan^{-1} x d x \). We set \( u = \tan^{-1} x \) and \( dv = 1 dx \). This gives \( du = \frac{1}{1+x^2} dx \) and \( v = x \). Applying the formula yields \( x \tan^{-1} x - \int x \frac{1}{1+x^2} dx \). For the remaining integral, we use a substitution \( t = 1+x^2 \), which results in \( \frac{1}{2} \log (1+x^2) \). Combining these and multiplying by 2 (from the initial simplification), the final answer is \( 2x \tan^{-1} x - \log (1+x^2) + C \). This problem highlights the importance of recognizing inverse trigonometric identities.
In simple words: We first use a special rule to change the complicated \( \tan^{-1} \) expression into \( 2 \tan^{-1} x \). Then, we use the integration by parts rule. After all the steps, the final answer is \( 2x \tan^{-1} x - \log (1+x^2) + C \).

🎯 Exam Tip: Memorize important inverse trigonometric identities, especially those like \( 2 \tan^{-1} x = \tan^{-1} \left(\frac{2x}{1-x^2}\right) \), \( \sin^{-1} \left(\frac{2x}{1+x^2}\right) \), or \( \cos^{-1} \left(\frac{1-x^2}{1+x^2}\right) \), as they simplify integrals significantly.

Question 29. \( \int \tan ^{-1}\left(\frac{2 x}{1-x^2}\right) d x \)
Answer: Let the integral be \( I \). We have \( I = \int \tan ^{-1}\left(\frac{2 x}{1-x^2}\right) d x \).
To simplify, let's use a substitution: let \( x = \tan \theta \). This means \( \theta = \tan^{-1} x \).
Next, find the differential \( dx \). Differentiating \( x = \tan \theta \) with respect to \( \theta \) gives \( dx = \sec^2 \theta \, d\theta \).
Substitute these into the integral:
\( I = \int \tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan^2 \theta}\right) \sec^2 \theta \, d\theta \)
We know the trigonometric identity \( \frac{2 \tan \theta}{1-\tan^2 \theta} = \tan (2\theta) \). So, the integral becomes:
\( I = \int \tan ^{-1}(\tan 2\theta) \sec^2 \theta \, d\theta \)
\( I = \int (2\theta) \sec^2 \theta \, d\theta \)
Now, we will use integration by parts, which is \( \int u \, dv = uv - \int v \, du \).
Let \( u = 2\theta \) and \( dv = \sec^2 \theta \, d\theta \).
Then \( du = 2 \, d\theta \) and \( v = \int \sec^2 \theta \, d\theta = \tan \theta \).
So,
\( I = 2\theta \tan \theta - \int \tan \theta (2 \, d\theta) \)
\( I = 2\theta \tan \theta - 2 \int \tan \theta \, d\theta \)
We know \( \int \tan \theta \, d\theta = \log |\sec \theta| + C \).
So,
\( I = 2\theta \tan \theta - 2 \log |\sec \theta| + C \)
Now, we need to substitute back \( x = \tan \theta \). We have \( \theta = \tan^{-1} x \).
From \( x = \tan \theta \), we can draw a right-angled triangle. If the opposite side is \( x \) and the adjacent side is \( 1 \), then the hypotenuse is \( \sqrt{1^2 + x^2} = \sqrt{1+x^2} \).
So, \( \sec \theta = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{\sqrt{1+x^2}}{1} = \sqrt{1+x^2} \).
Substitute these values back into the expression for \( I \):
\( I = 2(\tan^{-1} x)(x) - 2 \log |\sqrt{1+x^2}| + C \)
We can also write \( \log |\sqrt{1+x^2}| \) as \( \log (1+x^2)^{1/2} = \frac{1}{2} \log (1+x^2) \) using logarithm properties.
So,
\( I = 2x \tan^{-1} x - 2 \left(\frac{1}{2} \log (1+x^2)\right) + C \)
\( I = 2x \tan^{-1} x - \log (1+x^2) + C \)
This integral involves a standard form that can be solved using substitution and integration by parts. The substitution simplifies the inverse trigonometric function, making the integration straightforward.
In simple words: First, we change the variable from \( x \) to \( \theta \) using \( x = \tan \theta \). This makes the inverse tangent part much simpler. Then, we use a method called "integration by parts" to solve the new integral. Finally, we change \( \theta \) back to \( x \) to get the final answer.

🎯 Exam Tip: When you see \( \tan^{-1}\left(\frac{2x}{1-x^2}\right) \), always think of the substitution \( x = \tan \theta \) to simplify the argument of the inverse tangent function, as it relates to the double angle identity for tangent.

Question 30. \( \int \cos ^{-1} \frac{1-x^2}{1+x^2} d x \)
Answer: Let the integral be \( I \). We have \( I = \int \cos ^{-1} \frac{1-x^2}{1+x^2} d x \).
This expression \( \frac{1-x^2}{1+x^2} \) suggests another trigonometric substitution. Let's try \( x = \tan \theta \).
This means \( \theta = \tan^{-1} x \).
Then, \( dx = \sec^2 \theta \, d\theta \).
Substitute these into the integral:
\( I = \int \cos ^{-1}\left(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\right) \sec^2 \theta \, d\theta \)
We know the trigonometric identity \( \frac{1-\tan^2 \theta}{1+\tan^2 \theta} = \cos (2\theta) \). So, the integral becomes:
\( I = \int \cos ^{-1}(\cos 2\theta) \sec^2 \theta \, d\theta \)
\( I = \int (2\theta) \sec^2 \theta \, d\theta \)
This is the exact same integral as in Question 29. We can solve it using integration by parts.
Let \( u = 2\theta \) and \( dv = \sec^2 \theta \, d\theta \).
Then \( du = 2 \, d\theta \) and \( v = \int \sec^2 \theta \, d\theta = \tan \theta \).
Applying the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\( I = 2\theta \tan \theta - \int \tan \theta (2 \, d\theta) \)
\( I = 2\theta \tan \theta - 2 \int \tan \theta \, d\theta \)
\( I = 2\theta \tan \theta - 2 \log |\sec \theta| + C \)
Now, we convert back to \( x \). We have \( x = \tan \theta \), so \( \theta = \tan^{-1} x \).
From the right-angled triangle (as in Q29), if the opposite side is \( x \) and the adjacent side is \( 1 \), the hypotenuse is \( \sqrt{1+x^2} \).
So, \( \sec \theta = \sqrt{1+x^2} \).
Substitute these back:
\( I = 2(\tan^{-1} x)(x) - 2 \log |\sqrt{1+x^2}| + C \)
Using logarithm properties, \( \log |\sqrt{1+x^2}| = \frac{1}{2} \log (1+x^2) \).
\( I = 2x \tan^{-1} x - 2 \left(\frac{1}{2} \log (1+x^2)\right) + C \)
\( I = 2x \tan^{-1} x - \log (1+x^2) + C \)
The choice of substitution makes this complex integral much simpler to solve. It is a very common technique in integration.
In simple words: This problem looks tricky because of the inverse cosine function. But if we let \( x \) be tangent of an angle, the fraction inside the cosine becomes much simpler, turning into \( \cos(2\theta) \). After this, the problem becomes the same as the previous one, which we solve using integration by parts. Finally, we convert everything back to \( x \).

🎯 Exam Tip: Recognize the form \( \frac{1-x^2}{1+x^2} \) as an indicator for the substitution \( x = \tan \theta \), which leads to \( \cos(2\theta) \), simplifying the inverse cosine. This is a crucial trick in integration.

Question 31. \( \int\left(\tan ^{-1} x^2\right) x d x \)
Answer: Let the integral be \( I \). We have \( I = \int x \tan^{-1} (x^2) d x \).
This integral can be solved using substitution followed by integration by parts.
First, let's substitute \( t = x^2 \).
Then, differentiate \( t \) with respect to \( x \): \( dt = 2x \, dx \).
This means \( x \, dx = \frac{1}{2} dt \).
Substitute these into the integral:
\( I = \int \tan^{-1} (t) \cdot \frac{1}{2} dt \)
\( I = \frac{1}{2} \int \tan^{-1} (t) \, dt \)
Now, we need to integrate \( \tan^{-1} (t) \) with respect to \( t \). We will use integration by parts, treating \( \tan^{-1} (t) \) as \( \tan^{-1} (t) \cdot 1 \).
The integration by parts formula is \( \int u \, dv = uv - \int v \, du \).
Let \( u = \tan^{-1} (t) \) and \( dv = 1 \, dt \).
Then, \( du = \frac{1}{1+t^2} dt \) and \( v = \int 1 \, dt = t \).
Substitute these into the integration by parts formula:
\( \int \tan^{-1} (t) \, dt = t \tan^{-1} (t) - \int t \cdot \frac{1}{1+t^2} dt \)
\( \int \tan^{-1} (t) \, dt = t \tan^{-1} (t) - \int \frac{t}{1+t^2} dt \)
For the integral \( \int \frac{t}{1+t^2} dt \), we can use another simple substitution. Let \( w = 1+t^2 \). Then \( dw = 2t \, dt \), so \( t \, dt = \frac{1}{2} dw \).
\( \int \frac{t}{1+t^2} dt = \int \frac{1}{w} \cdot \frac{1}{2} dw = \frac{1}{2} \int \frac{1}{w} dw = \frac{1}{2} \log |w| + C' \)
Substitute \( w \) back: \( \frac{1}{2} \log |1+t^2| + C' \). Since \( 1+t^2 \) is always positive, we can write \( \frac{1}{2} \log (1+t^2) \).
So,
\( \int \tan^{-1} (t) \, dt = t \tan^{-1} (t) - \frac{1}{2} \log (1+t^2) + C' \)
Now, substitute this back into the expression for \( I \):
\( I = \frac{1}{2} \left( t \tan^{-1} (t) - \frac{1}{2} \log (1+t^2) \right) + C \)
\( I = \frac{1}{2} t \tan^{-1} (t) - \frac{1}{4} \log (1+t^2) + C \)
Finally, substitute back \( t = x^2 \):
\( I = \frac{1}{2} x^2 \tan^{-1} (x^2) - \frac{1}{4} \log (1+(x^2)^2) + C \)
\( I = \frac{1}{2} x^2 \tan^{-1} (x^2) - \frac{1}{4} \log (1+x^4) + C \)
The process involves a clever substitution to simplify the argument of the inverse tangent, followed by standard integration by parts. Remember to substitute back all variables to their original form.
In simple words: First, we replace \( x^2 \) with a new variable to make the problem easier. This changes the integral into one involving just \( \tan^{-1} (t) \). Then, we use a method called "integration by parts" to solve this simpler integral. In the end, we swap back the original variable to get the final answer.

🎯 Exam Tip: For integrals of the form \( \int x f(x^2) dx \), always try the substitution \( t = x^2 \) first, as it often simplifies the problem significantly before further integration techniques are applied.

Question 32. \( \int \sec ^{-1} \sqrt{x} dx \)
Answer: Let the integral be \( I \). We have \( I = \int \sec^{-1} \sqrt{x} dx \).
This integral requires a substitution to simplify the argument of the inverse secant function.
Let \( t = \sqrt{x} \).
Then, square both sides: \( t^2 = x \).
Differentiate \( x \) with respect to \( t \): \( dx = 2t \, dt \).
Substitute these into the integral:
\( I = \int \sec^{-1} (t) \cdot (2t \, dt) \)
\( I = 2 \int t \sec^{-1} (t) \, dt \)
Now, we need to use integration by parts for \( \int t \sec^{-1} (t) \, dt \). The integration by parts formula is \( \int u \, dv = uv - \int v \, du \).
According to ILATE rule (Inverse, Log, Algebraic, Trig, Exponential), \( \sec^{-1} (t) \) comes before \( t \).
Let \( u = \sec^{-1} (t) \) and \( dv = t \, dt \).
Then, \( du = \frac{1}{|t| \sqrt{t^2-1}} dt \) and \( v = \int t \, dt = \frac{t^2}{2} \). Since \( t=\sqrt{x} \) and \( x \) is usually positive, we can assume \( t>0 \), so \( |t|=t \).
Substitute these into the integration by parts formula:
\( \int t \sec^{-1} (t) \, dt = \sec^{-1} (t) \cdot \frac{t^2}{2} - \int \frac{t^2}{2} \cdot \frac{1}{t \sqrt{t^2-1}} dt \)
\( \int t \sec^{-1} (t) \, dt = \frac{t^2}{2} \sec^{-1} (t) - \frac{1}{2} \int \frac{t}{\sqrt{t^2-1}} dt \)
For the integral \( \int \frac{t}{\sqrt{t^2-1}} dt \), let \( w = t^2-1 \). Then \( dw = 2t \, dt \), so \( t \, dt = \frac{1}{2} dw \).
\( \int \frac{t}{\sqrt{t^2-1}} dt = \int \frac{1}{\sqrt{w}} \cdot \frac{1}{2} dw = \frac{1}{2} \int w^{-1/2} dw = \frac{1}{2} \cdot \frac{w^{1/2}}{1/2} + C' = w^{1/2} + C' = \sqrt{t^2-1} + C' \).
So,
\( \int t \sec^{-1} (t) \, dt = \frac{t^2}{2} \sec^{-1} (t) - \frac{1}{2} \sqrt{t^2-1} + C' \)
Now, multiply by 2 (from the initial \( 2 \int \) ):
\( I = 2 \left( \frac{t^2}{2} \sec^{-1} (t) - \frac{1}{2} \sqrt{t^2-1} \right) + C \)
\( I = t^2 \sec^{-1} (t) - \sqrt{t^2-1} + C \)
Finally, substitute back \( t = \sqrt{x} \). This means \( t^2 = x \).
\( I = x \sec^{-1} (\sqrt{x}) - \sqrt{x-1} + C \)
This integral requires both a substitution and integration by parts. Correctly identifying \( u \) and \( dv \) in integration by parts is key, especially when dealing with inverse trigonometric functions. The final step is to convert all variables back to \( x \).
In simple words: We first change \( \sqrt{x} \) to a new variable \( t \). This makes the integral easier to work with. Then, we use the "integration by parts" method, which helps us integrate functions that are products of two different types. After doing that, we change \( t \) back to \( \sqrt{x} \) to get our final answer.

🎯 Exam Tip: When integrating inverse trigonometric functions with a root in the argument (like \( \sqrt{x} \)), a substitution like \( t = \sqrt{x} \) or \( t^2 = x \) is often the first step to simplify the expression before applying integration by parts.

Question 33. \( \int \frac{\log \left(\sec ^{-1} x\right)}{x \sqrt{\left(x^2-1\right)}} d x \)
Answer: Let the integral be \( I \). We have \( I = \int \frac{\log \left(\sec^{-1} x\right)}{x \sqrt{x^2-1}} d x \).
This integral looks complex, but it can be simplified significantly with a substitution.
Notice the derivative of \( \sec^{-1} x \) is \( \frac{1}{x \sqrt{x^2-1}} \). This is a strong hint for substitution.
Let \( t = \sec^{-1} x \).
Then, differentiate \( t \) with respect to \( x \): \( dt = \frac{1}{x \sqrt{x^2-1}} dx \).
Substitute these into the integral:
\( I = \int \log(t) \, dt \)
Now, we need to integrate \( \log(t) \) with respect to \( t \). We will use integration by parts, treating \( \log(t) \) as \( \log(t) \cdot 1 \).
The integration by parts formula is \( \int u \, dv = uv - \int v \, du \).
Let \( u = \log(t) \) and \( dv = 1 \, dt \).
Then, \( du = \frac{1}{t} dt \) and \( v = \int 1 \, dt = t \).
Substitute these into the integration by parts formula:
\( \int \log(t) \, dt = t \log(t) - \int t \cdot \frac{1}{t} dt \)
\( \int \log(t) \, dt = t \log(t) - \int 1 \, dt \)
\( \int \log(t) \, dt = t \log(t) - t + C \)
Finally, substitute back \( t = \sec^{-1} x \):
\( I = (\sec^{-1} x) \log(\sec^{-1} x) - \sec^{-1} x + C \)
We can also factor out \( \sec^{-1} x \):
\( I = \sec^{-1} x (\log(\sec^{-1} x) - 1) + C \)
The key to solving this integral is recognizing the derivative of the inverse secant function within the integrand, which makes the initial substitution very powerful. This simplifies the entire problem to a standard integral of \( \log(t) \).
In simple words: We notice that a big part of the fraction is actually the derivative of \( \sec^{-1} x \). So, we replace \( \sec^{-1} x \) with a new variable \( t \). This turns the entire complex integral into a simple one: \( \int \log(t) \, dt \). We solve this using integration by parts, and then put \( \sec^{-1} x \) back in for \( t \) to get the final answer.

🎯 Exam Tip: Always look for derivatives of functions present in the integrand, especially with inverse trigonometric or logarithmic functions, as this often points to a direct substitution that greatly simplifies the integral.

Question 34. \( \int e^x(1+x) \log \left(x e^x\right) d x \)
Answer: Let the integral be \( I \). We have \( I = \int e^x(1+x) \log \left(x e^x\right) d x \).
This integral has a structure that suggests a substitution related to the product rule for differentiation.
Let's consider the function inside the logarithm: \( x e^x \).
Let \( t = x e^x \).
Now, differentiate \( t \) with respect to \( x \) using the product rule \( (uv)' = u'v + uv' \):
\( \frac{dt}{dx} = (1)e^x + x(e^x) = e^x + x e^x = e^x(1+x) \)
So, \( dt = e^x(1+x) \, dx \).
Notice that the term \( e^x(1+x) \, dx \) is exactly what we have in the integral outside the logarithm.
Substitute these into the integral:
\( I = \int \log(t) \, dt \)
This is the same integral as in Question 33. We integrate \( \log(t) \) using integration by parts.
Let \( u = \log(t) \) and \( dv = 1 \, dt \).
Then, \( du = \frac{1}{t} dt \) and \( v = t \).
Applying integration by parts:
\( I = t \log(t) - \int t \cdot \frac{1}{t} dt \)
\( I = t \log(t) - \int 1 \, dt \)
\( I = t \log(t) - t + C \)
Finally, substitute back \( t = x e^x \):
\( I = (x e^x) \log(x e^x) - x e^x + C \)
We can also factor out \( x e^x \):
\( I = x e^x (\log(x e^x) - 1) + C \)
This integral is a perfect example of how recognizing a function and its derivative can simplify a seemingly complex problem into a standard form. The specific form \( e^x(f(x) + f'(x)) \) often suggests a product rule in reverse.
In simple words: We look closely at the problem and see that if we consider \( x e^x \) as a new variable, its derivative \( e^x(1+x) \) is also present in the integral. So, we make this substitution, which changes the whole problem into a simple integral of \( \log(t) \). After solving that, we put \( x e^x \) back in place of \( t \) to get the final answer.

🎯 Exam Tip: When you see \( e^x \) multiplied by an expression like \( (1+x) \) and also a \( \log \) function, consider if the term in the log, \( f(x) \), has its derivative, \( f'(x) \), combined with \( e^x \) in the form \( e^x(f(x)+f'(x)) \). This often allows for a direct substitution using the product rule in reverse.

Question 35. \( \int \sin ^{-1}\left(3 x-4 x^3\right) d x \)
Answer: Let the integral be \( I \). We have \( I = \int \sin^{-1}(3x-4x^3) dx \).
The expression \( 3x-4x^3 \) is a familiar trigonometric identity. It resembles \( \sin(3\theta) \).
Let's use the substitution \( x = \sin \theta \).
This implies \( \theta = \sin^{-1} x \).
Then, differentiate \( x \) with respect to \( \theta \): \( dx = \cos \theta \, d\theta \).
Substitute these into the integral:
\( I = \int \sin^{-1}(3\sin\theta - 4\sin^3\theta) \cos\theta \, d\theta \)
Using the triple angle identity \( \sin(3\theta) = 3\sin\theta - 4\sin^3\theta \), the integral becomes:
\( I = \int \sin^{-1}(\sin(3\theta)) \cos\theta \, d\theta \)
\( I = \int (3\theta) \cos\theta \, d\theta \)
Now, we need to use integration by parts for \( 3 \int \theta \cos\theta \, d\theta \). The integration by parts formula is \( \int u \, dv = uv - \int v \, du \).
Let \( u = \theta \) and \( dv = \cos\theta \, d\theta \).
Then, \( du = d\theta \) and \( v = \int \cos\theta \, d\theta = \sin\theta \).
Substitute these into the integration by parts formula:
\( \int \theta \cos\theta \, d\theta = \theta \sin\theta - \int \sin\theta \, d\theta \)
\( \int \theta \cos\theta \, d\theta = \theta \sin\theta - (-\cos\theta) + C' \)
\( \int \theta \cos\theta \, d\theta = \theta \sin\theta + \cos\theta + C' \)
Now, multiply by 3 (from \( 3 \int \) ):
\( I = 3(\theta \sin\theta + \cos\theta) + C \)
Finally, substitute back \( \theta = \sin^{-1} x \). We also have \( x = \sin\theta \).
To find \( \cos\theta \) in terms of \( x \), use the identity \( \cos\theta = \sqrt{1-\sin^2\theta} = \sqrt{1-x^2} \). (We assume \( \theta \) is in the principal value range for \( \sin^{-1} x \), so \( \cos\theta \ge 0 \)).
Substitute these back into the expression for \( I \):
\( I = 3(x \sin^{-1} x + \sqrt{1-x^2}) + C \)
This problem leverages a key trigonometric identity to simplify the inverse sine function, transforming a complex integral into a manageable integration by parts problem. Knowing these identities is crucial for simplifying such integrals.
In simple words: The expression inside the \( \sin^{-1} \) function is a special one, which is actually \( \sin(3\theta) \) if we replace \( x \) with \( \sin\theta \). So, we use this substitution to simplify the integral to \( \int 3\theta \cos\theta \, d\theta \). Then we solve this using "integration by parts" and finally convert everything back to \( x \) using the original substitution and right-angled triangle properties.

🎯 Exam Tip: Memorize trigonometric identities like \( 3\sin\theta - 4\sin^3\theta = \sin(3\theta) \) and \( 4\cos^3\theta - 3\cos\theta = \cos(3\theta) \). They are frequently used to simplify inverse trigonometric integrals via substitution, making them solvable by integration by parts.

Question 36. \( \int \frac{x^2 \tan ^{-1} x}{1+x^2} d x \)
Answer: Let the integral be \( I \). We have \( I = \int \frac{x^2 \tan^{-1} x}{1+x^2} d x \).
This integral can be solved using substitution followed by integration by parts.
First, let's make a substitution for the inverse tangent function.
Let \( \theta = \tan^{-1} x \).
This implies \( x = \tan \theta \).
Then, differentiate \( x \) with respect to \( \theta \): \( dx = \sec^2 \theta \, d\theta \).
Substitute these into the integral:
\( I = \int \frac{(\tan \theta)^2 \cdot \theta}{1+(\tan \theta)^2} \sec^2 \theta \, d\theta \)
We know the identity \( 1+\tan^2 \theta = \sec^2 \theta \). So, the denominator \( (1+\tan^2 \theta) \) cancels out with \( \sec^2 \theta \) from \( dx \).
\( I = \int \frac{\tan^2 \theta \cdot \theta}{\sec^2 \theta} \sec^2 \theta \, d\theta \)
\( I = \int \theta \tan^2 \theta \, d\theta \)
We know another identity: \( \tan^2 \theta = \sec^2 \theta - 1 \).
So,
\( I = \int \theta (\sec^2 \theta - 1) \, d\theta \)
\( I = \int (\theta \sec^2 \theta - \theta) \, d\theta \)
We can split this into two integrals:
\( I = \int \theta \sec^2 \theta \, d\theta - \int \theta \, d\theta \)
The second integral is straightforward: \( \int \theta \, d\theta = \frac{\theta^2}{2} + C' \).
For the first integral, \( \int \theta \sec^2 \theta \, d\theta \), we use integration by parts. Let \( u = \theta \) and \( dv = \sec^2 \theta \, d\theta \).
Then, \( du = d\theta \) and \( v = \int \sec^2 \theta \, d\theta = \tan \theta \).
Applying integration by parts:
\( \int \theta \sec^2 \theta \, d\theta = \theta \tan \theta - \int \tan \theta \, d\theta \)
\( \int \theta \sec^2 \theta \, d\theta = \theta \tan \theta - \log |\sec \theta| + C'' \)
Combine both parts of \( I \):
\( I = (\theta \tan \theta - \log |\sec \theta|) - \frac{\theta^2}{2} + C \)
Now, substitute back \( \theta = \tan^{-1} x \) and \( x = \tan \theta \).
Also, \( \sec \theta = \sqrt{1+\tan^2 \theta} = \sqrt{1+x^2} \).
So,
\( I = (\tan^{-1} x) x - \log |\sqrt{1+x^2}| - \frac{(\tan^{-1} x)^2}{2} + C \)
We can write \( \log |\sqrt{1+x^2}| = \frac{1}{2} \log (1+x^2) \).
\( I = x \tan^{-1} x - \frac{1}{2} \log (1+x^2) - \frac{1}{2} (\tan^{-1} x)^2 + C \)
This integral first requires a substitution to change the variable and simplify the fraction, then uses integration by parts, and finally back-substitutes to the original variable. It's a multi-step process that combines several common integration techniques.
In simple words: We start by replacing \( \tan^{-1} x \) with an angle \( \theta \). This changes the whole integral into a simpler form involving \( \theta \) and \( \tan^2 \theta \). Then, we break it into two parts: one we solve directly and the other using "integration by parts." Finally, we replace \( \theta \) back with \( \tan^{-1} x \) to get the solution.

🎯 Exam Tip: For integrals containing \( \tan^{-1} x \) and \( \frac{x^2}{1+x^2} \), a substitution of \( \theta = \tan^{-1} x \) is usually very effective. Remember to simplify \( \frac{x^2}{1+x^2} \) using trigonometric identities like \( \tan^2 \theta = \sec^2 \theta - 1 \) after substitution.