RBSE Solutions Class 12 Maths Chapter 9 Integration Exercise 9.6

 

Question 1. Exalnate the following:
(a) \( \int x \cos x \, dx \)
(b) \( \int x \sec^2 x \, dx \)
Answer:
(a) To solve \( \int x \cos x \, dx \), we use integration by parts, which follows the formula \( \int u \, dv = uv - \int v \, du \). We choose \( u = x \) and \( dv = \cos x \, dx \).
\( \implies du = dx \) and \( v = \int \cos x \, dx = \sin x \)
Now, we plug these into the formula:
\( \int x \cos x \, dx = x \sin x - \int \sin x \, dx \)
\( = x \sin x - (-\cos x) + C \)
\( = x \sin x + \cos x + C \)
(b) To solve \( \int x \sec^2 x \, dx \), we again use integration by parts. This method helps solve integrals of products of functions that simplify or remain manageable after differentiation or integration.
We choose \( u = x \) and \( dv = \sec^2 x \, dx \).
\( \implies du = dx \) and \( v = \int \sec^2 x \, dx = \tan x \)
Plugging these into the formula:
\( \int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx \)
\( = x \tan x - (-\log |\cos x|) + C \)
\( = x \tan x + \log |\cos x| + C \)
This can also be written as:
\( = x \tan x - \log |\sec x| + C \)
In simple words: For both parts, we used a special rule called integration by parts. This rule helps us integrate when we have two different types of functions multiplied together. We pick one function to differentiate and the other to integrate, then follow the steps to find the final answer.

🎯 Exam Tip: Remember the LIATE rule (Logarithmic, Inverse, Algebraic, Trigonometric, Exponential) to choose 'u' for integration by parts. The function that appears earlier in LIATE is typically chosen as 'u'.

 

Question 2.
(a) \( \int x^3 e^{-x} \, dx \)
(b) \( \int x^3 \sin x \, dx \)
Answer:
(a) To solve \( \int x^3 e^{-x} \, dx \), we apply integration by parts multiple times. The formula is \( \int u \, dv = uv - \int v \, du \).
Step 1: Let \( u = x^3 \) and \( dv = e^{-x} \, dx \).
\( \implies du = 3x^2 \, dx \) and \( v = \int e^{-x} \, dx = -e^{-x} \)
\( \int x^3 e^{-x} \, dx = -x^3 e^{-x} - \int (-e^{-x}) 3x^2 \, dx = -x^3 e^{-x} + 3 \int x^2 e^{-x} \, dx \)
Step 2: For \( \int x^2 e^{-x} \, dx \), let \( u = x^2 \) and \( dv = e^{-x} \, dx \).
\( \implies du = 2x \, dx \) and \( v = -e^{-x} \)
\( \int x^2 e^{-x} \, dx = -x^2 e^{-x} - \int (-e^{-x}) 2x \, dx = -x^2 e^{-x} + 2 \int x e^{-x} \, dx \)
Step 3: For \( \int x e^{-x} \, dx \), let \( u = x \) and \( dv = e^{-x} \, dx \).
\( \implies du = dx \) and \( v = -e^{-x} \)
\( \int x e^{-x} \, dx = -x e^{-x} - \int (-e^{-x}) \, dx = -x e^{-x} + \int e^{-x} \, dx = -x e^{-x} - e^{-x} \)
Now, substitute back:
\( \int x^2 e^{-x} \, dx = -x^2 e^{-x} + 2 (-x e^{-x} - e^{-x}) = -x^2 e^{-x} - 2x e^{-x} - 2e^{-x} \)
Finally, substitute all back into the original integral:
\( \int x^3 e^{-x} \, dx = -x^3 e^{-x} + 3 (-x^2 e^{-x} - 2x e^{-x} - 2e^{-x}) + C \)
\( = -x^3 e^{-x} - 3x^2 e^{-x} - 6x e^{-x} - 6e^{-x} + C \)
\( = -e^{-x} (x^3 + 3x^2 + 6x + 6) + C \)
(b) To solve \( \int x^3 \sin x \, dx \), we also apply integration by parts repeatedly. This pattern of repeated integration by parts is common when one function (like a polynomial) eventually becomes zero after repeated differentiation.
Step 1: Let \( u = x^3 \) and \( dv = \sin x \, dx \).
\( \implies du = 3x^2 \, dx \) and \( v = \int \sin x \, dx = -\cos x \)
\( \int x^3 \sin x \, dx = -x^3 \cos x - \int (-\cos x) 3x^2 \, dx = -x^3 \cos x + 3 \int x^2 \cos x \, dx \)
Step 2: For \( \int x^2 \cos x \, dx \), let \( u = x^2 \) and \( dv = \cos x \, dx \).
\( \implies du = 2x \, dx \) and \( v = \int \cos x \, dx = \sin x \)
\( \int x^2 \cos x \, dx = x^2 \sin x - \int \sin x (2x) \, dx = x^2 \sin x - 2 \int x \sin x \, dx \)
Step 3: For \( \int x \sin x \, dx \), let \( u = x \) and \( dv = \sin x \, dx \).
\( \implies du = dx \) and \( v = -\cos x \)
\( \int x \sin x \, dx = -x \cos x - \int (-\cos x) \, dx = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x \)
Now, substitute back:
\( \int x^2 \cos x \, dx = x^2 \sin x - 2 (-x \cos x + \sin x) = x^2 \sin x + 2x \cos x - 2 \sin x \)
Finally, substitute all back into the original integral:
\( \int x^3 \sin x \, dx = -x^3 \cos x + 3 (x^2 \sin x + 2x \cos x - 2 \sin x) + C \)
\( = -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x + C \)
In simple words: For both problems, we had to use the integration by parts rule many times. We kept applying the rule until the 'x' term disappeared. For part (a), we integrated 'x cubed e to the power minus x', and for part (b), we integrated 'x cubed sine x'. It's a step-by-step method to break down complex integrals.

🎯 Exam Tip: Be extremely careful with signs when integrating by parts, especially with trigonometric and exponential functions involving negative powers. Tabular integration can be a helpful shortcut for these types of problems.

 

Question 3.
(a) \( \int x^3 (\log x)^2 \, dx \)
(b) \( \int x^3 e^{x^2} \, dx \)
Answer:
(a) To solve \( \int x^3 (\log x)^2 \, dx \), we use integration by parts. When a logarithmic term is present, it is often chosen as 'u' because its derivative is simpler.
Step 1: Let \( u = (\log x)^2 \) and \( dv = x^3 \, dx \).
\( \implies du = 2 \log x \cdot \frac{1}{x} \, dx \) and \( v = \int x^3 \, dx = \frac{x^4}{4} \)
\( \int x^3 (\log x)^2 \, dx = \frac{x^4}{4} (\log x)^2 - \int \frac{x^4}{4} \cdot 2 \log x \cdot \frac{1}{x} \, dx \)
\( = \frac{x^4}{4} (\log x)^2 - \frac{1}{2} \int x^3 \log x \, dx \)
Step 2: Now, for the integral \( \int x^3 \log x \, dx \), we apply integration by parts again.
Let \( u = \log x \) and \( dv = x^3 \, dx \).
\( \implies du = \frac{1}{x} \, dx \) and \( v = \frac{x^4}{4} \)
\( \int x^3 \log x \, dx = \log x \frac{x^4}{4} - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx \)
\( = \frac{x^4}{4} \log x - \frac{1}{4} \int x^3 \, dx \)
\( = \frac{x^4}{4} \log x - \frac{1}{4} \frac{x^4}{4} + C_1 = \frac{x^4}{4} \log x - \frac{x^4}{16} + C_1 \)
Substitute this back into the expression from Step 1:
\( \int x^3 (\log x)^2 \, dx = \frac{x^4}{4} (\log x)^2 - \frac{1}{2} \left( \frac{x^4}{4} \log x - \frac{x^4}{16} \right) + C \)
\( = \frac{x^4}{4} (\log x)^2 - \frac{x^4}{8} \log x + \frac{x^4}{32} + C \)
This can also be written as:
\( = \frac{x^4}{4} \left[ (\log x)^2 - \frac{1}{2} \log x + \frac{1}{8} \right] + C \)
(b) To solve \( \int x^3 e^{x^2} \, dx \), we first use a substitution. Substitution is a powerful technique that simplifies many complex integrals by transforming them into a more recognizable form.
Let \( t = x^2 \). Then \( dt = 2x \, dx \implies x \, dx = \frac{1}{2} dt \).
Rewrite \( x^3 e^{x^2} \, dx \) as \( x^2 e^{x^2} \cdot x \, dx \).
Now substitute:
\( \int x^2 e^{x^2} \cdot x \, dx = \int t e^t \frac{1}{2} dt = \frac{1}{2} \int t e^t \, dt \)
Next, we use integration by parts for \( \int t e^t \, dt \).
Let \( u = t \) and \( dv = e^t \, dt \).
\( \implies du = dt \) and \( v = \int e^t \, dt = e^t \)
\( \int t e^t \, dt = t e^t - \int e^t \, dt = t e^t - e^t \)
Substitute this back:
\( \frac{1}{2} \int t e^t \, dt = \frac{1}{2} (t e^t - e^t) + C \)
\( = \frac{1}{2} e^t (t-1) + C \)
Finally, substitute \( t = x^2 \) back into the expression:
\( = \frac{1}{2} e^{x^2} (x^2-1) + C \)
In simple words: For part (a), we needed to integrate 'x cubed' with 'log x squared'. We used the integration by parts rule two times because of the 'log x squared' term. For part (b), we first made a simple change, letting 'x squared' be 't'. This made the integral easier to handle, turning it into a form we could solve with integration by parts, and then we changed 't' back to 'x squared'.

🎯 Exam Tip: For integrals involving \( (\log x)^n \), repeated integration by parts is usually required. For integrals of the form \( \int x^{2n+1} e^{x^2} \, dx \) or similar, a substitution like \( t = x^2 \) often simplifies the problem significantly, leading to a standard integration by parts form.

 

Question 4.
(a) \( \int e^{2x} e^{e^x} \, dx \)
(b) \( \int (\log x)^2 \, dx \)
Answer:
(a) No solution is provided in the source content for this part of the question.
(b) To solve \( \int (\log x)^2 \, dx \), we use integration by parts, taking '1' as the second function. Integrating \( (\log x)^n \) often involves taking '1' as the second function for integration by parts.
Step 1: Let \( u = (\log x)^2 \) and \( dv = 1 \, dx \).
\( \implies du = 2 \log x \cdot \frac{1}{x} \, dx \) and \( v = \int 1 \, dx = x \)
\( \int (\log x)^2 \, dx = x (\log x)^2 - \int x \cdot \left( 2 \log x \cdot \frac{1}{x} \right) \, dx \)
\( = x (\log x)^2 - 2 \int \log x \, dx \)
Step 2: Now, for the integral \( \int \log x \, dx \), we apply integration by parts again, with \( u = \log x \) and \( dv = 1 \, dx \).
\( \implies du = \frac{1}{x} \, dx \) and \( v = x \)
\( \int \log x \, dx = x \log x - \int x \cdot \frac{1}{x} \, dx \)
\( = x \log x - \int 1 \, dx = x \log x - x \)
Substitute this back into the expression from Step 1:
\( \int (\log x)^2 \, dx = x (\log x)^2 - 2 (x \log x - x) + C \)
\( = x (\log x)^2 - 2x \log x + 2x + C \)
In simple words: For part (b), we want to integrate 'log x squared'. We do this by pretending there's a '1' multiplied with it, and then use the integration by parts rule two times. This helps us get rid of the logarithm and find the complete answer.

🎯 Exam Tip: When integrating inverse trigonometric or logarithmic functions, always consider '1' as the second function for integration by parts if no other suitable function is present.

 

Question 5.
(a) \( \int \cos^{-1} x \, dx \)
(b) \( \int \csc^{-1} \sqrt{\frac{x+a}{x}} \, dx \)
Answer:
(a) To solve \( \int \cos^{-1} x \, dx \), we use integration by parts with \( u = \cos^{-1} x \) and \( dv = 1 \, dx \).
\( \implies du = \frac{-1}{\sqrt{1-x^2}} \, dx \) and \( v = x \)
\( \int \cos^{-1} x \, dx = x \cos^{-1} x - \int x \cdot \left( \frac{-1}{\sqrt{1-x^2}} \right) \, dx \)
\( = x \cos^{-1} x + \int \frac{x}{\sqrt{1-x^2}} \, dx \)
For the integral \( \int \frac{x}{\sqrt{1-x^2}} \, dx \), we use a substitution. The derivative of inverse trigonometric functions often contains a square root term, which can sometimes be integrated using a simple substitution.
Let \( t = 1-x^2 \). Then \( dt = -2x \, dx \implies x \, dx = -\frac{1}{2} dt \).
\( \int \frac{x}{\sqrt{1-x^2}} \, dx = \int \frac{-\frac{1}{2} dt}{\sqrt{t}} = -\frac{1}{2} \int t^{-1/2} \, dt \)
\( = -\frac{1}{2} \frac{t^{1/2}}{1/2} + C_1 = -\sqrt{t} + C_1 = -\sqrt{1-x^2} + C_1 \)
Substitute this back:
\( \int \cos^{-1} x \, dx = x \cos^{-1} x - \sqrt{1-x^2} + C \)
(b) To solve \( \int \csc^{-1} \sqrt{\frac{x+a}{x}} \, dx \), we first simplify the inverse cosecant function and use a trigonometric substitution. Trigonometric substitutions are very effective for simplifying integrals involving square roots of algebraic expressions.
We know \( \csc^{-1} y = \sin^{-1} \left( \frac{1}{y} \right) \), so \( \csc^{-1} \sqrt{\frac{x+a}{x}} = \sin^{-1} \left( \frac{1}{\sqrt{\frac{x+a}{x}}} \right) = \sin^{-1} \sqrt{\frac{x}{x+a}} \).
Let \( x = a \tan^2 \theta \).
Then \( dx = \frac{d}{d\theta}(a \tan^2 \theta) \, d\theta = 2a \tan \theta \sec^2 \theta \, d\theta \).
Now, simplify \( \sqrt{\frac{x}{x+a}} \):
\( \sqrt{\frac{x}{x+a}} = \sqrt{\frac{a \tan^2 \theta}{a \tan^2 \theta + a}} = \sqrt{\frac{a \tan^2 \theta}{a(1+\tan^2 \theta)}} = \sqrt{\frac{\tan^2 \theta}{\sec^2 \theta}} = \sqrt{\sin^2 \theta} = \sin \theta \)
So, \( \sin^{-1} \sqrt{\frac{x}{x+a}} = \sin^{-1} (\sin \theta) = \theta \).
The integral becomes \( \int \theta \cdot (2a \tan \theta \sec^2 \theta) \, d\theta = 2a \int \theta \tan \theta \sec^2 \theta \, d\theta \).
Now, use integration by parts with \( u = \theta \) and \( dv = 2 \tan \theta \sec^2 \theta \, d\theta \).
\( \implies du = d\theta \). To find \( v \): Let \( w = \tan \theta \), then \( dw = \sec^2 \theta \, d\theta \).
\( v = \int 2 \tan \theta \sec^2 \theta \, d\theta = \int 2w \, dw = w^2 = \tan^2 \theta \).
Applying integration by parts:
\( 2a \left[ \theta \tan^2 \theta - \int \tan^2 \theta \, d\theta \right] \)
We know \( \int \tan^2 \theta \, d\theta = \int (\sec^2 \theta - 1) \, d\theta = \tan \theta - \theta \).
So, \( 2a \left[ \frac{\theta \tan^2 \theta}{2} - \frac{1}{2} (\tan \theta - \theta) \right] + C \) (The OCR has a factor of 'a' instead of '2a' at the start of solution, implying \( a \int \theta (2 \tan \theta \sec^2 \theta) d\theta \) where \( \int 2 \tan \theta \sec^2 \theta d\theta = \tan^2 \theta \). I will follow the OCR's steps for the integral expression.)
\( = a \left[ \theta \tan^2 \theta - \int \tan^2 \theta \, d\theta \right] \)
\( = a \left[ \theta \tan^2 \theta - (\tan \theta - \theta) \right] + C \)
\( = a \left[ \theta \tan^2 \theta - \tan \theta + \theta \right] + C \)
\( = a \left[ \theta (1+\tan^2 \theta) - \tan \theta \right] + C \)
(To express this back in terms of x: \( \theta = \tan^{-1} \left( \sqrt{\frac{x}{a}} \right) \), \( \tan \theta = \sqrt{\frac{x}{a}} \), and \( 1+\tan^2 \theta = 1+\frac{x}{a} = \frac{a+x}{a} \).)
\( = a \left[ \tan^{-1} \left( \sqrt{\frac{x}{a}} \right) \left( \frac{a+x}{a} \right) - \sqrt{\frac{x}{a}} \right] + C \)
\( = (x+a) \tan^{-1} \left( \sqrt{\frac{x}{a}} \right) - \sqrt{ax} + C \)
In simple words: For part (a), we integrated 'inverse cosine x' using a rule called integration by parts. Then, we solved the remaining part using a simple change of variables. For part (b), we first changed the 'inverse cosecant' into 'inverse sine' and then used a special trick: we replaced 'x' with 'a times tan squared theta'. This made the integral much simpler, and we solved it using integration by parts with 'theta' and 'tan squared theta' before changing back to 'x'.

🎯 Exam Tip: For inverse trigonometric integrals, consider integration by parts with '1' as the second function. For integrals involving square roots of algebraic expressions, look for trigonometric substitutions (e.g., \( x=a \tan^2 \theta \)) to simplify the terms before integrating.

 

Question 6.
(a) \( \int \sin^{-1} (3x-4x^3) \, dx \)
(b) \( \int \frac{x}{1+\cos x} \, dx \)
Answer:
(a) To solve \( \int \sin^{-1} (3x-4x^3) \, dx \), we use a trigonometric identity to simplify the term inside the inverse sine function. Recognizing trigonometric identities can drastically simplify complex inverse trigonometric expressions within integrals.
We know the identity: \( \sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta \).
Let \( x = \sin t \). Then \( dx = \cos t \, dt \).
The expression \( 3x - 4x^3 \) becomes \( 3 \sin t - 4 \sin^3 t = \sin 3t \).
So, the integral simplifies to:
\( \int \sin^{-1} (\sin 3t) \cos t \, dt = \int 3t \cos t \, dt \)
Now, we use integration by parts for \( \int 3t \cos t \, dt \).
Let \( u = 3t \) and \( dv = \cos t \, dt \).
\( \implies du = 3 \, dt \) and \( v = \int \cos t \, dt = \sin t \)
Applying the integration by parts formula:
\( \int 3t \cos t \, dt = 3t \sin t - \int \sin t \cdot 3 \, dt \)
\( = 3t \sin t - 3 \int \sin t \, dt \)
\( = 3t \sin t - 3 (-\cos t) + C \)
\( = 3t \sin t + 3 \cos t + C \)
Finally, substitute back \( t = \sin^{-1} x \).
Since \( x = \sin t \), we have \( \sin t = x \).
Also, \( \cos t = \sqrt{1-\sin^2 t} = \sqrt{1-x^2} \).
So, the integral is:
\( = 3x \sin^{-1} x + 3 \sqrt{1-x^2} + C \)
(b) To solve \( \int \frac{x}{1+\cos x} \, dx \), we first use a trigonometric identity to simplify the denominator. Trigonometric identities are often the first step to simplify integrands before applying integration techniques.
We use the half-angle formula \( 1+\cos x = 2 \cos^2 \frac{x}{2} \).
The integral becomes:
\( \int \frac{x}{2 \cos^2 \frac{x}{2}} \, dx = \int \frac{1}{2} x \sec^2 \frac{x}{2} \, dx \)
Now, we use integration by parts for \( \int \frac{1}{2} x \sec^2 \frac{x}{2} \, dx \).
Let \( u = x \) and \( dv = \frac{1}{2} \sec^2 \frac{x}{2} \, dx \).
\( \implies du = dx \). To find \( v \), we integrate \( \frac{1}{2} \sec^2 \frac{x}{2} \). Let \( w = \frac{x}{2} \), so \( dw = \frac{1}{2} dx \).
\( v = \int \sec^2 w \, dw = \tan w = \tan \frac{x}{2} \)
Applying the integration by parts formula:
\( \int \frac{1}{2} x \sec^2 \frac{x}{2} \, dx = x \tan \frac{x}{2} - \int \tan \frac{x}{2} \, dx \)
For the integral \( \int \tan \frac{x}{2} \, dx \), let \( y = \frac{x}{2} \), so \( dy = \frac{1}{2} dx \implies dx = 2 \, dy \).
\( \int \tan \frac{x}{2} \, dx = \int \tan y \cdot 2 \, dy = 2 \int \tan y \, dy = 2 \log |\sec y| + C_1 = 2 \log \left|\sec \frac{x}{2}\right| + C_1 \)
Substitute this back:
\( = x \tan \frac{x}{2} - 2 \log \left|\sec \frac{x}{2}\right| + C \)
In simple words: For part (a), we noticed a special trigonometry pattern inside the inverse sine function. By changing 'x' to 'sine t', the expression became 'sine 3t', which helped simplify the integral greatly. We then solved this simpler integral using the integration by parts rule. For part (b), we used a half-angle identity for '1 plus cos x' to make the bottom part simpler. This allowed us to apply integration by parts to the new expression, simplifying it step by step.

🎯 Exam Tip: Always look for trigonometric identities that can simplify the integrand before applying integration by parts or other complex integration techniques. This often makes the problem much more manageable.

 

Question 7.
(a) \( \int \tan^{-1} \sqrt{\frac{1-x}{1+x}} \, dx \)
(b) \( \int \cos \sqrt{x} \, dx \)
Answer:
(a) To solve \( \int \tan^{-1} \sqrt{\frac{1-x}{1+x}} \, dx \), we use a trigonometric substitution to simplify the expression inside the inverse tangent. Using trigonometric substitutions like \( x = \cos \theta \) is a standard technique to simplify expressions involving \( \sqrt{\frac{1-x}{1+x}} \).
Let \( x = \cos \theta \). Then \( dx = -\sin \theta \, d\theta \).
Now, simplify the term inside the square root:
\( \sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} = \sqrt{\frac{2 \sin^2 (\theta/2)}{2 \cos^2 (\theta/2)}} = \sqrt{\tan^2 (\theta/2)} = \tan (\theta/2) \)
So, \( \tan^{-1} \sqrt{\frac{1-x}{1+x}} = \tan^{-1} (\tan (\theta/2)) = \frac{\theta}{2} \).
The integral becomes:
\( \int \frac{\theta}{2} (-\sin \theta) \, d\theta = -\frac{1}{2} \int \theta \sin \theta \, d\theta \)
Now, we use integration by parts for \( \int \theta \sin \theta \, d\theta \).
Let \( u = \theta \) and \( dv = \sin \theta \, d\theta \).
\( \implies du = d\theta \) and \( v = \int \sin \theta \, d\theta = -\cos \theta \)
Applying the integration by parts formula:
\( \int \theta \sin \theta \, d\theta = \theta (-\cos \theta) - \int (-\cos \theta) \, d\theta \)
\( = -\theta \cos \theta + \int \cos \theta \, d\theta \)
\( = -\theta \cos \theta + \sin \theta \)
Substitute this back into the main integral expression:
\( -\frac{1}{2} [-\theta \cos \theta + \sin \theta] + C \)
\( = \frac{1}{2} \theta \cos \theta - \frac{1}{2} \sin \theta + C \)
Finally, substitute back \( \theta = \cos^{-1} x \).
Since \( x = \cos \theta \), we have \( \cos \theta = x \).
Also, \( \sin \theta = \sqrt{1-\cos^2 \theta} = \sqrt{1-x^2} \).
So, the integral is:
\( = \frac{1}{2} x \cos^{-1} x - \frac{1}{2} \sqrt{1-x^2} + C \)
(b) To solve \( \int \cos \sqrt{x} \, dx \), we start with a substitution to simplify the argument of the cosine function. A substitution that removes a square root from the argument of a trigonometric function can often lead to a standard integration by parts problem.
Let \( \sqrt{x} = t \). Then squaring both sides gives \( x = t^2 \).
Now, differentiate to find \( dx \): \( dx = 2t \, dt \).
Substitute these into the integral:
\( \int \cos \sqrt{x} \, dx = \int \cos t \cdot (2t \, dt) = 2 \int t \cos t \, dt \)
Now, we use integration by parts for \( \int t \cos t \, dt \).
Let \( u = t \) and \( dv = \cos t \, dt \).
\( \implies du = dt \) and \( v = \int \cos t \, dt = \sin t \)
Applying the integration by parts formula:
\( \int t \cos t \, dt = t \sin t - \int \sin t \, dt \)
\( = t \sin t - (-\cos t) + C_1 \)
\( = t \sin t + \cos t + C_1 \)
Substitute this back into the main integral expression:
\( 2 \int t \cos t \, dt = 2 [t \sin t + \cos t] + C \)
Finally, substitute back \( t = \sqrt{x} \):
\( = 2 [\sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}] + C \)
In simple words: For part (a), we replaced 'x' with 'cos theta'. This made the complex expression inside the inverse tangent much simpler, turning it into 'theta over 2'. We then used integration by parts to solve this easier integral and changed 'theta' back to 'x'. For part (b), we first changed 'square root x' to 't'. This transformed the integral into '2t cos t', which we solved using integration by parts, and then changed 't' back to 'square root x'.

🎯 Exam Tip: When the argument of an inverse trigonometric function involves \( \sqrt{\frac{1-x}{1+x}} \), try the substitution \( x = \cos \theta \). For integrals like \( \int f(\sqrt{x}) \, dx \), substituting \( \sqrt{x} = t \) (or \( x = t^2 \)) is often the key first step.

 

Question 9. \( \int x \sin^{-1} x / \sqrt{1-x^2} \, dx \)
Answer: Let the integral be \( I \). We start by using a substitution to simplify it.
Let \( t = \sin^{-1} x \).
Then \( x = \sin t \).
Now, find the differential \( dx \): \( dx = \cos t \, dt \).
Substitute these into the integral:
\( I = \int \sin t \cdot t \cdot \frac{1}{\sqrt{1 - \sin^2 t}} \cdot \cos t \, dt \)
We know that \( \sqrt{1 - \sin^2 t} = \sqrt{\cos^2 t} = \cos t \). So,
\( I = \int \sin t \cdot t \cdot \frac{1}{\cos t} \cdot \cos t \, dt \)
\( I = \int t \sin t \, dt \)
Now, we use integration by parts, which states \( \int u \, dv = uv - \int v \, du \).
Let \( u = t \) and \( dv = \sin t \, dt \).
Then \( du = dt \) and \( v = \int \sin t \, dt = -\cos t \).
Substitute these into the integration by parts formula:
\( I = t(-\cos t) - \int (-\cos t) \, dt \)
\( I = -t \cos t + \int \cos t \, dt \)
\( I = -t \cos t + \sin t + C \)
Finally, substitute back \( t = \sin^{-1} x \) and \( \sin t = x \). We also need \( \cos t \). Since \( \sin t = x \), we can use the identity \( \cos t = \sqrt{1 - \sin^2 t} = \sqrt{1 - x^2} \).
\( I = -\sin^{-1} x \cdot \sqrt{1 - x^2} + x + C \)
In simple words: To solve this, we first swap \( \sin^{-1} x \) for a new letter, say \( t \). This changes the whole problem into a simpler one. After doing the math, we put the original letters back to get the final answer. This method simplifies complex integrals.

🎯 Exam Tip: For integrals involving inverse trigonometric functions, substitution with the inverse function is often the first step to convert it into a standard form that can be solved with integration by parts.

 

Question 10. \( \int \frac{x \tan^{-1} x}{(1+x^2)^{3/2}} \, dx \)
Answer: Let the integral be \( I \). To make this integral easier, we use a substitution.
Let \( x = \tan \theta \).
Then, the differential \( dx = \sec^2 \theta \, d\theta \).
Now, substitute \( x \) and \( dx \) into the integral:
\( I = \int \frac{\tan \theta \cdot \tan^{-1}(\tan \theta)}{(1 + \tan^2 \theta)^{3/2}} \sec^2 \theta \, d\theta \)
We know that \( \tan^{-1}(\tan \theta) = \theta \) and \( 1 + \tan^2 \theta = \sec^2 \theta \). So, the expression becomes:
\( I = \int \frac{\tan \theta \cdot \theta}{(\sec^2 \theta)^{3/2}} \sec^2 \theta \, d\theta \)
\( I = \int \frac{\theta \tan \theta}{(\sec^3 \theta)} \sec^2 \theta \, d\theta \)
\( I = \int \frac{\theta \tan \theta}{\sec \theta} \, d\theta \)
Since \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \frac{1}{\sec \theta} = \cos \theta \):
\( I = \int \theta \left( \frac{\sin \theta}{\cos \theta} \right) \cos \theta \, d\theta \)
\( I = \int \theta \sin \theta \, d\theta \)
Now, we use integration by parts. Let \( u = \theta \) and \( dv = \sin \theta \, d\theta \).
Then \( du = d\theta \) and \( v = \int \sin \theta \, d\theta = -\cos \theta \).
Applying the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\( I = \theta(-\cos \theta) - \int (-\cos \theta) \, d\theta \)
\( I = -\theta \cos \theta + \int \cos \theta \, d\theta \)
\( I = -\theta \cos \theta + \sin \theta + C \)
Now, we need to convert back from \( \theta \) to \( x \). Since \( x = \tan \theta \), we have \( \theta = \tan^{-1} x \).
Also, from \( x = \tan \theta \), we can draw a right triangle where the opposite side is \( x \) and the adjacent side is 1. The hypotenuse is \( \sqrt{x^2+1} \).
So, \( \sin \theta = \frac{x}{\sqrt{x^2+1}} \) and \( \cos \theta = \frac{1}{\sqrt{x^2+1}} \).
Substitute these values back into the expression for \( I \):
\( I = -\tan^{-1} x \cdot \frac{1}{\sqrt{x^2+1}} + \frac{x}{\sqrt{x^2+1}} + C \)
This can be written as:
\( I = \frac{x - \tan^{-1} x}{\sqrt{x^2+1}} + C \)
In simple words: We changed \( x \) to \( \tan \theta \) to make the integral simpler. After doing some steps, we used a rule called 'integration by parts'. Finally, we changed everything back to \( x \) to get the solution. This process helps us solve integrals that look complicated at first.

🎯 Exam Tip: When you see \( 1+x^2 \) or \( \sqrt{1+x^2} \) in an integral, consider substituting \( x=\tan \theta \) or \( x=\cot \theta \) as it often simplifies the expression using trigonometric identities.

 

Question 12. \( \int \frac{2x + \sin 2x}{1 + \cos 2x} \, dx \)
Answer: Let the integral be \( I \). We start by using trigonometric identities to simplify the integrand.
We know that \( \sin 2x = 2 \sin x \cos x \) and \( 1 + \cos 2x = 2 \cos^2 x \).
Substitute these identities into the integral:
\( I = \int \frac{2x + 2 \sin x \cos x}{2 \cos^2 x} \, dx \)
Now, separate the fraction into two parts:
\( I = \int \left( \frac{2x}{2 \cos^2 x} + \frac{2 \sin x \cos x}{2 \cos^2 x} \right) \, dx \)
Simplify each term:
\( I = \int \left( x \cdot \frac{1}{\cos^2 x} + \frac{\sin x}{\cos x} \right) \, dx \)
\( I = \int (x \sec^2 x + \tan x) \, dx \)
We can split this into two separate integrals:
\( I = \int x \sec^2 x \, dx + \int \tan x \, dx \)
For the first integral, \( \int x \sec^2 x \, dx \), we use integration by parts. Let \( u = x \) and \( dv = \sec^2 x \, dx \).
Then \( du = dx \) and \( v = \int \sec^2 x \, dx = \tan x \).
Applying the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\( \int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx \)
Now, substitute this back into the expression for \( I \):
\( I = (x \tan x - \int \tan x \, dx) + \int \tan x \, dx \)
The \( -\int \tan x \, dx \) and \( +\int \tan x \, dx \) terms cancel each other out.
\( I = x \tan x + C \)
In simple words: We first used known math rules for \( \sin 2x \) and \( \cos 2x \) to make the problem look simpler. Then we split the problem into two parts. For one part, we used a special rule called 'integration by parts'. Many parts cancelled out, leaving us with a simple answer.

🎯 Exam Tip: Always look for opportunities to simplify trigonometric expressions using identities before attempting integration. This can drastically reduce the complexity of the problem.

 

Question 13. \( \int e^x \left( \frac{1-\sin x}{1-\cos x} \right) \, dx \)
Answer: Let the integral be \( I \). We will use half-angle trigonometric formulas to simplify the expression inside the integral.
We know that \( 1 - \cos x = 2 \sin^2(x/2) \) and \( \sin x = 2 \sin(x/2) \cos(x/2) \).
Substitute these identities into the integral:
\( I = \int e^x \left( \frac{1 - 2 \sin(x/2) \cos(x/2)}{2 \sin^2(x/2)} \right) \, dx \)
Now, separate the fraction into two terms:
\( I = \int e^x \left( \frac{1}{2 \sin^2(x/2)} - \frac{2 \sin(x/2) \cos(x/2)}{2 \sin^2(x/2)} \right) \, dx \)
Simplify each term using \( \frac{1}{\sin^2(x/2)} = \operatorname{cosec}^2(x/2) \) and \( \frac{\cos(x/2)}{\sin(x/2)} = \cot(x/2) \):
\( I = \int e^x \left( \frac{1}{2} \operatorname{cosec}^2(x/2) - \cot(x/2) \right) \, dx \)
This integral is of a special form: \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \).
Here, let \( f(x) = -\cot(x/2) \).
Then, find the derivative of \( f(x) \):
\( f'(x) = - \left( -\operatorname{cosec}^2(x/2) \right) \cdot \frac{d}{dx}(x/2) \)
\( f'(x) = \operatorname{cosec}^2(x/2) \cdot \frac{1}{2} = \frac{1}{2} \operatorname{cosec}^2(x/2) \)
Since the integrand is exactly in the form \( e^x [f(x) + f'(x)] \), we can directly write the result:
\( I = e^x \cdot (-\cot(x/2)) + C \)
\( I = -e^x \cot(x/2) + C \)
In simple words: We used special angle rules to make the messy part of the integral simpler. This changed the problem into a known pattern where \( e^x \) is multiplied by a function and its derivative. Once it matched the pattern, the answer became very easy to find.

🎯 Exam Tip: Always look for the \( \int e^x [f(x) + f'(x)] \, dx \) pattern when \( e^x \) is present. Knowing half-angle formulas for \( 1-\cos x \) and \( \sin x \) is crucial for simplifying such problems.

 

Question 14. \( \int e^x \left[ \log x + \frac{1}{x^2} \right] \, dx \)
Answer: Let the integral be \( I \). The provided solution implicitly solves for a slightly different integral structure, which is a common form in integration by parts.
The solution steps provided in the source indicate that the problem is being treated as \( \int e^x \left( \frac{1}{x^2} - \frac{2}{x^3} \right) \, dx \). We will proceed with the method shown in the source, which correctly applies to this latter form.
We are solving \( I = \int e^x \left( \frac{1}{x^2} - \frac{2}{x^3} \right) \, dx \)
This integral is of the form \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \).
Let \( f(x) = \frac{1}{x^2} = x^{-2} \).
Now, find the derivative of \( f(x) \):
\( f'(x) = \frac{d}{dx} (x^{-2}) = -2x^{-2-1} = -2x^{-3} = -\frac{2}{x^3} \).
Since our integral is \( \int e^x \left( x^{-2} + (-2x^{-3}) \right) \, dx \), it exactly matches the form \( \int e^x [f(x) + f'(x)] \, dx \).
Therefore, the solution is directly:
\( I = e^x \cdot f(x) + C \)
\( I = e^x \cdot \frac{1}{x^2} + C \)
\( I = \frac{e^x}{x^2} + C \)
In simple words: This problem uses a special rule for integrals with \( e^x \). If you can spot a function and its derivative inside the brackets with \( e^x \), the answer is simply \( e^x \) times that function. Here, \( 1/x^2 \) is the function, and \( -2/x^3 \) is its derivative, making the solution straightforward.

🎯 Exam Tip: Memorize the pattern \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \). This saves a lot of time by avoiding integration by parts when the integral fits this form. Identify \( f(x) \) first and then check if the other term is its derivative.

 

Question 15. \( \int e^x [\log (\sec x + \tan x) + \sec x] \, dx \)
Answer: Let the integral be \( I \). This integral is in a recognizable form for easy solution.
The integral is of the form \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \).
Here, let \( f(x) = \log(\sec x + \tan x) \).
Now, we need to find the derivative of \( f(x) \) to check if it matches the other term in the integral.
\( f'(x) = \frac{d}{dx} [\log(\sec x + \tan x)] \)
Using the chain rule, \( \frac{d}{dx} \log(u) = \frac{1}{u} \frac{du}{dx} \):
\( f'(x) = \frac{1}{\sec x + \tan x} \cdot \frac{d}{dx} (\sec x + \tan x) \)
\( f'(x) = \frac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x) \)
Factor out \( \sec x \) from the numerator:
\( f'(x) = \frac{\sec x (\tan x + \sec x)}{\sec x + \tan x} \)
The term \( (\tan x + \sec x) \) cancels out from the numerator and denominator.
\( f'(x) = \sec x \)
Since the integral is exactly in the form \( \int e^x [f(x) + f'(x)] \, dx \) where \( f(x) = \log(\sec x + \tan x) \) and \( f'(x) = \sec x \), the solution is:
\( I = e^x f(x) + C \)
\( I = e^x \log(\sec x + \tan x) + C \)
In simple words: This integral follows a simple rule where if you have \( e^x \) multiplied by a function plus its derivative, the answer is just \( e^x \) times the original function. Here, \( \log(\sec x + \tan x) \) is the function, and \( \sec x \) is its derivative, making the solution quick.

🎯 Exam Tip: Recognizing the derivative of \( \log(\sec x + \tan x) \) as \( \sec x \) is a key trick. This identity is very useful in quickly solving integrals of the form \( \int e^x [f(x) + f'(x)] \, dx \).

 

Question 16. \( \int e^x (\sin x + \cos x) \sec^2 x \, dx \)
Answer: Let the integral be \( I \). We start by distributing \( \sec^2 x \) inside the parenthesis.
\( I = \int e^x (\sin x \sec^2 x + \cos x \sec^2 x) \, dx \)
Now, let's simplify the terms inside the parenthesis using \( \sec^2 x = \frac{1}{\cos^2 x} \):
\( \sin x \sec^2 x = \sin x \cdot \frac{1}{\cos^2 x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \tan x \sec x \)
\( \cos x \sec^2 x = \cos x \cdot \frac{1}{\cos^2 x} = \frac{1}{\cos x} = \sec x \)
Substitute these simplified terms back into the integral:
\( I = \int e^x (\tan x \sec x + \sec x) \, dx \)
Rearrange the terms inside the parenthesis:
\( I = \int e^x (\sec x + \sec x \tan x) \, dx \)
This integral is now in the special form \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \).
Here, let \( f(x) = \sec x \).
Then, find the derivative of \( f(x) \):
\( f'(x) = \frac{d}{dx} (\sec x) = \sec x \tan x \)
Since the integrand is exactly in the form \( e^x [f(x) + f'(x)] \), the solution is:
\( I = e^x f(x) + C \)
\( I = e^x \sec x + C \)
In simple words: First, we multiplied \( \sec^2 x \) by both parts inside the bracket and simplified them using basic trigonometry. This made the problem fit a special rule for integrals with \( e^x \). We found a function (\( \sec x \)) and its derivative (\( \sec x \tan x \)) within the integral, which allowed us to get the answer quickly.

🎯 Exam Tip: Always try to expand and simplify trigonometric expressions inside the integral, especially when \( e^x \) is a factor. This often reveals the \( e^x [f(x) + f'(x)] \) pattern, which is a significant shortcut.

 

Question 17. \( \int e^x \left( \frac{1}{x^2} - \frac{2}{x^3} \right) \, dx \)
Answer: Let the integral be \( I \). This integral is in a direct form for easy solution.
The integral is of the form \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \).
Here, let \( f(x) = \frac{1}{x^2} \). We can also write this as \( f(x) = x^{-2} \).
Now, find the derivative of \( f(x) \):
\( f'(x) = \frac{d}{dx} (x^{-2}) \)
Using the power rule \( \frac{d}{dx} (x^n) = nx^{n-1} \):
\( f'(x) = -2x^{-2-1} = -2x^{-3} \)
This can also be written as \( f'(x) = -\frac{2}{x^3} \).
The integrand in the question is \( e^x \left( \frac{1}{x^2} - \frac{2}{x^3} \right) \), which perfectly matches \( e^x [f(x) + f'(x)] \).
Therefore, the solution is:
\( I = e^x f(x) + C \)
\( I = e^x \cdot \frac{1}{x^2} + C \)
\( I = \frac{e^x}{x^2} + C \)
In simple words: This problem is solved using a special rule for integrals with \( e^x \). When the part inside the bracket is a function plus its derivative, the answer is just \( e^x \) multiplied by that function. Here, \( 1/x^2 \) is the function, and \( -2/x^3 \) is its derivative.

🎯 Exam Tip: Always look for the \( \int e^x [f(x) + f'(x)] \, dx \) pattern when \( e^x \) is present. Identifying \( f(x) \) and then quickly checking if the other term is its derivative will lead to a rapid solution.

 

Question 18. \( \int e^x \left( \frac{1-x}{1+x^2} \right)^2 \, dx \)
Answer: Let the integral be \( I \). We start by expanding the square in the numerator and denominator.
\( I = \int e^x \frac{(1-x)^2}{(1+x^2)^2} \, dx \)
Expand the numerator \( (1-x)^2 = 1 - 2x + x^2 \):
\( I = \int e^x \frac{1 - 2x + x^2}{(1+x^2)^2} \, dx \)
Now, we rearrange the numerator to find a pattern suitable for the \( \int e^x [f(x) + f'(x)] \, dx \) form.
We can write \( 1 + x^2 - 2x \) instead of \( 1 - 2x + x^2 \).
\( I = \int e^x \frac{(1+x^2) - 2x}{(1+x^2)^2} \, dx \)
Split the fraction into two terms:
\( I = \int e^x \left( \frac{1+x^2}{(1+x^2)^2} - \frac{2x}{(1+x^2)^2} \right) \, dx \)
Simplify the first term:
\( I = \int e^x \left( \frac{1}{1+x^2} - \frac{2x}{(1+x^2)^2} \right) \, dx \)
This integral is now in the special form \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \).
Here, let \( f(x) = \frac{1}{1+x^2} \).
Now, find the derivative of \( f(x) \):
\( f'(x) = \frac{d}{dx} \left( (1+x^2)^{-1} \right) \)
Using the chain rule:
\( f'(x) = -1 \cdot (1+x^2)^{-2} \cdot \frac{d}{dx}(1+x^2) \)
\( f'(x) = -1 \cdot (1+x^2)^{-2} \cdot (2x) = -\frac{2x}{(1+x^2)^2} \)
Since the integrand matches \( e^x [f(x) + f'(x)] \), the solution is:
\( I = e^x f(x) + C \)
\( I = \frac{e^x}{1+x^2} + C \)
In simple words: We first opened up the square in the top part of the fraction and then rearranged it. This helped us split the fraction into two simpler parts. One part turned out to be a function, and the other part was its derivative, which made the solution easy using a special rule for \( e^x \) integrals.

🎯 Exam Tip: For integrals involving \( e^x \) and fractions with \( (1+x^2) \), try to manipulate the numerator to create terms that match \( f(x) \) and \( f'(x) \) for \( f(x) = \frac{1}{1+x^2} \).

 

Question 19. \( \int \cos 2\theta \cdot \log \left( \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta} \right) \, d\theta \)
Answer: Let the integral be \( I \). We will use integration by parts for this problem.
First, let's simplify the logarithmic term. We know that \( \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta} = \frac{1 + \tan \theta}{1 - \tan \theta} = \tan\left(\frac{\pi}{4} + \theta\right) \).
So, let \( u = \log \left( \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta} \right) = \log(\tan(\frac{\pi}{4} + \theta)) \).
And let \( dv = \cos 2\theta \, d\theta \).
Now, we find \( du \) and \( v \).
To find \( du \), we differentiate \( u \):
\( du = \frac{1}{\tan(\frac{\pi}{4} + \theta)} \cdot \sec^2(\frac{\pi}{4} + \theta) \cdot 1 \, d\theta \)
\( du = \frac{\cos(\frac{\pi}{4} + \theta)}{\sin(\frac{\pi}{4} + \theta)} \cdot \frac{1}{\cos^2(\frac{\pi}{4} + \theta)} \, d\theta \)
\( du = \frac{1}{\sin(\frac{\pi}{4} + \theta) \cos(\frac{\pi}{4} + \theta)} \, d\theta \)
Multiply numerator and denominator by 2:
\( du = \frac{2}{2 \sin(\frac{\pi}{4} + \theta) \cos(\frac{\pi}{4} + \theta)} \, d\theta \)
Using the identity \( 2 \sin A \cos A = \sin 2A \):
\( du = \frac{2}{\sin(2(\frac{\pi}{4} + \theta))} \, d\theta = \frac{2}{\sin(\frac{\pi}{2} + 2\theta)} \, d\theta \)
Since \( \sin(\frac{\pi}{2} + X) = \cos X \):
\( du = \frac{2}{\cos 2\theta} \, d\theta = 2 \sec 2\theta \, d\theta \).
To find \( v \), we integrate \( dv \):
\( v = \int \cos 2\theta \, d\theta = \frac{\sin 2\theta}{2} \).
Now, apply the integration by parts formula: \( \int u \, dv = uv - \int v \, du \).
\( I = \left[ \log \left( \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta} \right) \right] \left( \frac{\sin 2\theta}{2} \right) - \int \left( \frac{\sin 2\theta}{2} \right) (2 \sec 2\theta) \, d\theta \)
\( I = \frac{\sin 2\theta}{2} \log \left( \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta} \right) - \int \frac{\sin 2\theta}{\cos 2\theta} \, d\theta \)
\( I = \frac{\sin 2\theta}{2} \log \left( \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta} \right) - \int \tan 2\theta \, d\theta \)
The integral of \( \tan 2\theta \) is \( \frac{1}{2} \log|\sec 2\theta| \). So, \( -\int \tan 2\theta \, d\theta = -\frac{1}{2} \log|\sec 2\theta| \).
This can also be written as \( \frac{1}{2} \log|\cos 2\theta| \) (since \( -\log|A| = \log|1/A| \)).
\( I = \frac{\sin 2\theta}{2} \log \left( \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta} \right) + \frac{1}{2} \log|\cos 2\theta| + C \)
In simple words: This integral needed a special technique called 'integration by parts'. We first made the logarithm term simpler by rewriting it using a tangent identity. Then, we found the derivative of one part and the integral of the other part. We put these into the integration by parts rule, which helped us solve the problem and get the final answer.

🎯 Exam Tip: When dealing with logarithmic terms in integrals, especially those involving trigonometric functions, try to simplify the logarithm's argument using trigonometric identities first. The derivative of \( \log(\tan(\frac{\pi}{4} + \theta)) \) often simplifies surprisingly well, which is key here.

 

Question 20. \( \int \frac{x^2}{(x \cos x - \sin x)^2} \, dx \)
Answer: Let the integral be \( I \). This integral requires a specific approach using integration by parts after a clever rewrite.
We can rewrite \( x^2 \) as \( (x \operatorname{cosec} x)(x \sin x) \). So the integral becomes:
\( I = \int (x \operatorname{cosec} x) \cdot \frac{x \sin x}{(x \cos x - \sin x)^2} \, dx \)
Now, we use integration by parts, \( \int U \, dV = UV - \int V \, dU \).
Let \( U = x \operatorname{cosec} x \).
Let \( dV = \frac{x \sin x}{(x \cos x - \sin x)^2} \, dx \).
First, let's find \( V \) by integrating \( dV \). We can do this using a substitution.
Let \( t = x \cos x - \sin x \).
Then \( dt = \frac{d}{dx}(x \cos x - \sin x) \, dx \).
\( dt = (\cos x - x \sin x - \cos x) \, dx \)
\( dt = -x \sin x \, dx \).
So, \( x \sin x \, dx = -dt \).
Now substitute into the integral for \( V \):
\( V = \int \frac{-dt}{t^2} = -\int t^{-2} \, dt = - \left( \frac{t^{-1}}{-1} \right) + C_1 = \frac{1}{t} + C_1 \)
Substituting back \( t = x \cos x - \sin x \):
\( V = \frac{1}{x \cos x - \sin x} \). (We omit \( C_1 \) for now and add it at the end).
Next, we find \( dU \) by differentiating \( U \):
\( dU = \frac{d}{dx} (x \operatorname{cosec} x) \, dx \)
Using the product rule \( (fg)' = f'g + fg' \):
\( dU = (1 \cdot \operatorname{cosec} x + x \cdot (-\operatorname{cosec} x \cot x)) \, dx \)
\( dU = (\operatorname{cosec} x - x \operatorname{cosec} x \cot x) \, dx \)
\( dU = \operatorname{cosec} x (1 - x \cot x) \, dx \).
Now, apply the integration by parts formula:
\( I = U V - \int V \, dU \)
\( I = (x \operatorname{cosec} x) \left( \frac{1}{x \cos x - \sin x} \right) - \int \left( \frac{1}{x \cos x - \sin x} \right) (\operatorname{cosec} x (1 - x \cot x)) \, dx \)
\( I = \frac{x \operatorname{cosec} x}{x \cos x - \sin x} - \int \frac{\operatorname{cosec} x (1 - x \cot x)}{x \cos x - \sin x} \, dx \)
Let's simplify the integrand of the second part:
\( \frac{\operatorname{cosec} x (1 - x \cot x)}{x \cos x - \sin x} = \frac{\frac{1}{\sin x} (1 - x \frac{\cos x}{\sin x})}{x \cos x - \sin x} \)
\( = \frac{\frac{1}{\sin x} \left( \frac{\sin x - x \cos x}{\sin x} \right)}{x \cos x - \sin x} = \frac{\sin x - x \cos x}{\sin^2 x (x \cos x - \sin x)} \)
\( = \frac{- (x \cos x - \sin x)}{\sin^2 x (x \cos x - \sin x)} = -\frac{1}{\sin^2 x} = -\operatorname{cosec}^2 x \).
So the integral becomes:
\( I = \frac{x \operatorname{cosec} x}{x \cos x - \sin x} - \int (-\operatorname{cosec}^2 x) \, dx \)
\( I = \frac{x \operatorname{cosec} x}{x \cos x - \sin x} + \int \operatorname{cosec}^2 x \, dx \)
We know \( \int \operatorname{cosec}^2 x \, dx = -\cot x \).
\( I = \frac{x \operatorname{cosec} x}{x \cos x - \sin x} - \cot x + C \)
Now, let's simplify the first term: \( \operatorname{cosec} x = \frac{1}{\sin x} \) and \( \cot x = \frac{\cos x}{\sin x} \).
\( I = \frac{x/\sin x}{x \cos x - \sin x} - \frac{\cos x}{\sin x} + C \)
\( I = \frac{x}{\sin x (x \cos x - \sin x)} - \frac{\cos x}{\sin x} + C \)
Combine the fractions:
\( I = \frac{x - \cos x (x \cos x - \sin x)}{\sin x (x \cos x - \sin x)} + C \)
\( I = \frac{x - x \cos^2 x + \sin x \cos x}{\sin x (x \cos x - \sin x)} + C \)
Factor out \( x \) from the first two terms:
\( I = \frac{x(1 - \cos^2 x) + \sin x \cos x}{\sin x (x \cos x - \sin x)} + C \)
Using \( 1 - \cos^2 x = \sin^2 x \):
\( I = \frac{x \sin^2 x + \sin x \cos x}{\sin x (x \cos x - \sin x)} + C \)
Factor out \( \sin x \) from the numerator:
\( I = \frac{\sin x (x \sin x + \cos x)}{\sin x (x \cos x - \sin x)} + C \)
Cancel \( \sin x \) from numerator and denominator:
\( I = \frac{x \sin x + \cos x}{x \cos x - \sin x} + C \)
In simple words: This integral was quite tricky and needed a clever first step to rewrite the top part. Then, we used a technique called 'integration by parts' and a substitution to break it down. After many steps of simplifying and combining terms, we arrived at the final answer.

🎯 Exam Tip: For complex fractions with powers, especially involving \( x \) and trigonometric functions, look for ways to rewrite the integrand to fit integration by parts. Identifying the derivative of the denominator (or a part of it) can simplify finding \( V \) in integration by parts.

 

Question 21. \( \int \cos^{-1} \left( \frac{1}{x} \right) \, dx \)
Answer: Let the integral be \( I \). We start by using a substitution involving the inverse cosine function.
Let \( t = \cos^{-1} \left( \frac{1}{x} \right) \).
Then, \( \cos t = \frac{1}{x} \).
This implies \( x = \sec t \).
Now, find the differential \( dx \): \( dx = \frac{d}{dt}(\sec t) \, dt = \sec t \tan t \, dt \).
Substitute \( t \) and \( dx \) into the integral:
\( I = \int t \cdot (\sec t \tan t) \, dt \)
Now, we use integration by parts. The formula is \( \int u \, dv = uv - \int v \, du \).
Let \( u = t \) and \( dv = \sec t \tan t \, dt \).
Then, \( du = dt \) and \( v = \int \sec t \tan t \, dt = \sec t \).
Apply the integration by parts formula:
\( I = t \sec t - \int \sec t \, dt \)
The integral of \( \sec t \) is \( \log|\sec t + \tan t| \).
\( I = t \sec t - \log|\sec t + \tan t| + C \)
Finally, substitute back to express the answer in terms of \( x \):
We have \( t = \cos^{-1} \left( \frac{1}{x} \right) \) and \( \sec t = x \).
To find \( \tan t \), we use the identity \( \tan t = \sqrt{\sec^2 t - 1} \).
\( \tan t = \sqrt{x^2 - 1} \).
Substitute these back into the expression for \( I \):
\( I = x \cos^{-1} \left( \frac{1}{x} \right) - \log|x + \sqrt{x^2 - 1}| + C \)
In simple words: We first changed the inverse cosine part to a new letter. This made the problem into a form that could be solved using the 'integration by parts' rule. After solving, we put the original variable back to get the final answer. This is a common method for integrals with inverse functions.

🎯 Exam Tip: When integrating inverse trigonometric functions, a substitution where the inverse function equals \( t \) is almost always the correct first step. This converts the integral into a simpler trigonometric form that is often solvable by parts.

 

Question 22. \( \int (\sin^{-1} x)^2 \, dx \)
Answer: Let the integral be \( I \). We start by using a substitution to simplify the inverse sine term.
Let \( t = \sin^{-1} x \).
Then, \( x = \sin t \).
Now, find the differential \( dx \): \( dx = \frac{d}{dt}(\sin t) \, dt = \cos t \, dt \).
Substitute \( t \) and \( dx \) into the integral:
\( I = \int t^2 \cos t \, dt \)
This integral requires integration by parts, which is \( \int u \, dv = uv - \int v \, du \). We will need to apply it twice.
**First application of integration by parts:**
Let \( u = t^2 \) and \( dv = \cos t \, dt \).
Then, \( du = 2t \, dt \) and \( v = \int \cos t \, dt = \sin t \).
Applying the formula:
\( I = t^2 \sin t - \int \sin t \cdot (2t) \, dt \)
\( I = t^2 \sin t - 2 \int t \sin t \, dt \)
**Second application of integration by parts (for \( \int t \sin t \, dt \)):**
Let \( U = t \) and \( dV = \sin t \, dt \).
Then, \( dU = dt \) and \( V = \int \sin t \, dt = -\cos t \).
Applying the formula for \( \int t \sin t \, dt \):
\( \int t \sin t \, dt = t(-\cos t) - \int (-\cos t) \, dt \)
\( \int t \sin t \, dt = -t \cos t + \int \cos t \, dt \)
\( \int t \sin t \, dt = -t \cos t + \sin t \).
Now, substitute this result back into the main integral expression for \( I \):
\( I = t^2 \sin t - 2(-t \cos t + \sin t) + C \)
\( I = t^2 \sin t + 2t \cos t - 2 \sin t + C \)
Finally, substitute back to express the answer in terms of \( x \):
We have \( t = \sin^{-1} x \) and \( \sin t = x \).
To find \( \cos t \), we use the identity \( \cos t = \sqrt{1 - \sin^2 t} \).
\( \cos t = \sqrt{1 - x^2} \).
Substitute these values back into the expression for \( I \):
\( I = (\sin^{-1} x)^2 \cdot x + 2(\sin^{-1} x) \cdot \sqrt{1 - x^2} - 2x + C \)
This can be written as:
\( I = x(\sin^{-1} x)^2 + 2\sqrt{1 - x^2} \sin^{-1} x - 2x + C \)
In simple words: We first replaced the inverse sine part with a new letter. This changed the problem into an integral that needed a rule called 'integration by parts' to be used twice. After solving it step-by-step, we put the original variable back to get the final answer. This method helps solve integrals involving squares of inverse functions.

🎯 Exam Tip: Integrals of the form \( \int (\text{inverse function})^2 \, dx \) almost always require a substitution for the inverse function, followed by two applications of integration by parts. Be careful with the signs and constants during each step of integration by parts.

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