RBSE Solutions Class 12 Maths Chapter 9 Integration Exercise 9.1

Question 1. Integrate the following with respect to x :
(a) \( 3x\sqrt{x^2} \)
(b) \( e^{3x} \)
(c) \( \left(\frac{1}{2}\right)^x \)
(d) \( a^2 \log_a x \)
Answer:
(a) First, simplify the expression: \( 3x\sqrt{x^2} = 3x \cdot x = 3x^2 \). Then, integrate.
\( \int 3x\sqrt{x^2} dx = \int 3x^2 dx \)
\( = 3 \frac{x^{2+1}}{2+1} + C \)
\( = 3 \frac{x^3}{3} + C \)
\( = x^3 + C \)
(b) For exponential functions of the form \( e^{kx} \), the integral is \( \frac{e^{kx}}{k} \).
\( \int e^{3x} dx = \frac{e^{3x}}{3} + C \)
(c) For exponential functions of the form \( a^x \), the integral is \( \frac{a^x}{\log_e a} \).
\( \int \left(\frac{1}{2}\right)^x dx = \frac{\left(\frac{1}{2}\right)^x}{\log_e \left(\frac{1}{2}\right)} + C \)
(d) The integration for \( a^2 \log_a x \) is shown by the source as if it were for \( x^2 \). We follow the provided steps.
\( \int a^2 \log_a x dx = \int x^2 dx \)
\( = \frac{x^3}{3} + C \)
In simple words: Integration helps us find the original function when we know its rate of change. We use specific rules for powers, exponential terms, and logarithms, always adding a constant 'C' at the end.

🎯 Exam Tip: Always remember to simplify the integrand before performing integration, as shown in part (a), and correctly apply the appropriate integration formula for each type of function.

Question 2. Find the value of the integral given below: \( \int \left(5 \cos x - 3 \sin x + \frac{2}{\cos^2 x}\right) dx \)
Answer: We can split the integral into three parts and simplify the third term using trigonometric identities.
\( \int \left(5 \cos x - 3 \sin x + \frac{2}{\cos^2 x}\right) dx \)
\( = \int 5 \cos x dx - \int 3 \sin x dx + \int 2 \sec^2 x dx \)
\( = 5 \int \cos x dx - 3 \int \sin x dx + 2 \int \sec^2 x dx \)
\( = 5 (\sin x) - 3 (-\cos x) + 2 (\tan x) + C \)
\( = 5 \sin x + 3 \cos x + 2 \tan x + C \)
In simple words: To solve this, break the big problem into smaller, easier parts. Integrate each part using standard formulas for cosine, sine, and secant squared, then combine the results.

🎯 Exam Tip: When integrating sums or differences of functions, integrate each term separately. Remember common integral formulas like \( \int \sin x dx = -\cos x \) and \( \int \sec^2 x dx = \tan x \).

Question 3. Integrate \( \frac{x^2-1}{x^2} \) with respect to x.
Answer: First, simplify the fraction by dividing each term in the numerator by the denominator. This makes the integration simpler.
\( \int \frac{x^2-1}{x^2} dx = \int \left(\frac{x^2}{x^2} - \frac{1}{x^2}\right) dx \)
\( = \int \left(1 - x^{-2}\right) dx \)
Now, integrate each term separately.
\( = \int 1 dx - \int x^{-2} dx \)
\( = x - \frac{x^{-2+1}}{-2+1} + C \)
\( = x - \frac{x^{-1}}{-1} + C \)
\( = x + x^{-1} + C \)
\( = x + \frac{1}{x} + C \)
In simple words: First, split the fraction into simpler parts. Then, integrate each part by itself, remembering to add 1 to the power and divide by the new power for terms like \( x^{-2} \).

🎯 Exam Tip: Always simplify complex fractions into a sum or difference of simpler terms before integrating. This often involves algebraic manipulation or splitting the numerator over the denominator.

Question 4. Integrate \( \int (\sec^2 x + \operatorname{cosec}^2 x) dx \).
Answer: We can integrate each trigonometric term separately using standard integration formulas. The integral of \( \sec^2 x \) is \( \tan x \) and the integral of \( \operatorname{cosec}^2 x \) is \( -\cot x \).
\( \int (\sec^2 x + \operatorname{cosec}^2 x) dx \)
\( = \int \sec^2 x dx + \int \operatorname{cosec}^2 x dx \)
\( = \tan x - \cot x + C \)
In simple words: We take the integral of secant squared, which is tangent x. Then we take the integral of cosecant squared, which is negative cotangent x. Put them together for the final answer.

🎯 Exam Tip: Memorize the basic integrals of common trigonometric functions like \( \sec^2 x \) and \( \operatorname{cosec}^2 x \). This helps in quickly solving direct integration problems.

Question 5. Integrate \( \int (1+x)\sqrt{x} dx \).
Answer: First, expand the expression by multiplying \( \sqrt{x} \) with each term inside the parenthesis. Then, convert the square roots to fractional exponents and integrate.
\( \int (1+x)\sqrt{x} dx = \int (1 \cdot x^{1/2} + x \cdot x^{1/2}) dx \)
\( = \int (x^{1/2} + x^{3/2}) dx \)
Now, integrate each term using the power rule for integration \( \int x^n dx = \frac{x^{n+1}}{n+1} \).
\( = \int x^{1/2} dx + \int x^{3/2} dx \)
\( = \frac{x^{1/2+1}}{1/2+1} + \frac{x^{3/2+1}}{3/2+1} + C \)
\( = \frac{x^{3/2}}{3/2} + \frac{x^{5/2}}{5/2} + C \)
\( = \frac{2}{3}x^{3/2} + \frac{2}{5}x^{5/2} + C \)
In simple words: Multiply out the terms first, changing square roots into powers like \( x^{1/2} \). Then, add 1 to each power and divide by that new power to integrate.

🎯 Exam Tip: Always expand or simplify algebraic expressions before integrating. Converting roots to fractional exponents simplifies the application of the power rule of integration.

Question 7. Integrate \( \int \frac{x^2}{1+x^2} dx \).
Answer: To integrate this, we can add and subtract 1 in the numerator to simplify the fraction. This trick helps separate the terms into easier integrals.
\( \int \frac{x^2}{1+x^2} dx = \int \frac{x^2+1-1}{1+x^2} dx \)
\( = \int \left(\frac{x^2+1}{1+x^2} - \frac{1}{1+x^2}\right) dx \)
\( = \int \left(1 - \frac{1}{1+x^2}\right) dx \)
Now, integrate each term separately.
\( = \int 1 dx - \int \frac{1}{1+x^2} dx \)
\( = x - \tan^{-1} x + C \)
In simple words: Make the top part look like the bottom part by adding and taking away 1. Then, split the fraction and integrate the simple terms, remembering the integral of \( 1/(1+x^2) \) is inverse tangent x.

🎯 Exam Tip: For rational functions where the degree of the numerator is equal to or greater than the degree of the denominator, perform polynomial long division or use algebraic manipulation (like adding and subtracting terms) to simplify the integrand before integrating.

Question 8. Integrate \( \int \frac{\cos^2 x}{1+\sin x} dx \).
Answer: Use the trigonometric identity \( \cos^2 x = 1 - \sin^2 x \) to rewrite the numerator. This will help simplify the fraction by factoring.
\( \int \frac{\cos^2 x}{1+\sin x} dx = \int \frac{1-\sin^2 x}{1+\sin x} dx \)
Now, use the difference of squares formula \( a^2-b^2 = (a-b)(a+b) \) in the numerator.
\( = \int \frac{(1-\sin x)(1+\sin x)}{1+\sin x} dx \)
Cancel out the common term \( (1+\sin x) \).
\( = \int (1-\sin x) dx \)
Integrate each term separately.
\( = \int 1 dx - \int \sin x dx \)
\( = x - (-\cos x) + C \)
\( = x + \cos x + C \)
In simple words: Change \( \cos^2 x \) to \( 1-\sin^2 x \). Then, split \( 1-\sin^2 x \) into two parts using a math rule and cancel out the matching part from the bottom. Finally, integrate the remaining simple terms.

🎯 Exam Tip: When dealing with trigonometric integrals, always look for opportunities to use fundamental identities (like \( \sin^2 x + \cos^2 x = 1 \)) to simplify the expression, especially before attempting direct integration.

Question 9. Integrate \( \int \sec x (\sec x + \tan x) dx \).
Answer: First, distribute \( \sec x \) into the parenthesis to expand the expression. This will give us two common integral forms.
\( \int \sec x (\sec x + \tan x) dx = \int (\sec^2 x + \sec x \tan x) dx \)
Now, integrate each term separately using their standard integral formulas.
\( = \int \sec^2 x dx + \int \sec x \tan x dx \)
\( = \tan x + \sec x + C \)
In simple words: Multiply \( \sec x \) by each term inside the bracket. This gives \( \sec^2 x \) and \( \sec x \tan x \). Then, use the simple rules to integrate each of these parts.

🎯 Exam Tip: Expanding the integrand (like multiplying terms) is often the first step in simplifying complex expressions into integrable forms. Always remember the standard integrals for \( \sec^2 x \) and \( \sec x \tan x \).

Question 10. Integrate \( \int (\sin^{-1} x + \cos^{-1} x) dx \).
Answer: We use a fundamental identity for inverse trigonometric functions which states that for any \( x \in [-1, 1] \), \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \). This simplifies the integrand to a constant.
\( \int (\sin^{-1} x + \cos^{-1} x) dx \)
\( = \int \frac{\pi}{2} dx \)
Now, integrate the constant with respect to x.
\( = \frac{\pi}{2} x + C \)
In simple words: There's a special rule that says \( \sin^{-1} x + \cos^{-1} x \) is always equal to \( \pi/2 \) (a constant number). So we just integrate this constant number, which means multiplying it by x.

🎯 Exam Tip: Always look for opportunities to simplify the integrand using identities, especially for inverse trigonometric functions. Recognizing \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \) is a key shortcut.

Question 11. Integrate \( \int \frac{x^2-1}{x^2+1} dx \).
Answer: To integrate this rational function, we can add and subtract 2 in the numerator to simplify it. This makes the numerator a multiple of the denominator minus a constant.
\( \int \frac{x^2-1}{x^2+1} dx = \int \frac{x^2+1-2}{x^2+1} dx \)
\( = \int \left(\frac{x^2+1}{x^2+1} - \frac{2}{x^2+1}\right) dx \)
\( = \int \left(1 - \frac{2}{x^2+1}\right) dx \)
Now, integrate each term separately.
\( = \int 1 dx - 2 \int \frac{1}{x^2+1} dx \)
\( = x - 2 \tan^{-1} x + C \)
In simple words: Change the top part by adding and taking away 2 so it looks more like the bottom part. This lets you split the fraction into simple parts that are easy to integrate.

🎯 Exam Tip: When integrating fractions where the numerator and denominator have similar terms, try adding and subtracting constants in the numerator to create terms that can be easily simplified or integrated into standard forms like \( \tan^{-1} x \).

Question 12. Integrate \( \int \tan^2 x dx \).
Answer: We cannot directly integrate \( \tan^2 x \), so we use the Pythagorean trigonometric identity: \( \tan^2 x = \sec^2 x - 1 \). This transforms the function into a form that is easy to integrate.
\( \int \tan^2 x dx = \int (\sec^2 x - 1) dx \)
Now, integrate each term separately.
\( = \int \sec^2 x dx - \int 1 dx \)
\( = \tan x - x + C \)
In simple words: You can't directly integrate \( \tan^2 x \). So, change it to \( \sec^2 x - 1 \) using a math rule. Then, integrate \( \sec^2 x \) to get \( \tan x \) and integrate 1 to get x.

🎯 Exam Tip: Always convert \( \tan^2 x \) into \( \sec^2 x - 1 \) for integration problems, as \( \sec^2 x \) is a standard integral (resulting in \( \tan x \)), while \( \tan^2 x \) is not directly integrable.

Question 13. Integrate \( \int \cot^2 x dx \).
Answer: Similar to \( \tan^2 x \), we cannot directly integrate \( \cot^2 x \). We use the trigonometric identity: \( \cot^2 x = \operatorname{cosec}^2 x - 1 \). This changes the function into easily integrable terms.
\( \int \cot^2 x dx = \int (\operatorname{cosec}^2 x - 1) dx \)
Now, integrate each term separately.
\( = \int \operatorname{cosec}^2 x dx - \int 1 dx \)
\( = -\cot x - x + C \)
In simple words: Since we can't integrate \( \cot^2 x \) directly, change it to \( \operatorname{cosec}^2 x - 1 \) using a math rule. Then, integrate \( \operatorname{cosec}^2 x \) to get \( -\cot x \) and integrate 1 to get x.

🎯 Exam Tip: Remember to convert \( \cot^2 x \) to \( \operatorname{cosec}^2 x - 1 \) when integrating, as \( \operatorname{cosec}^2 x \) is a standard integral (yielding \( -\cot x \)), making the integration possible.

Question 14. Integrate \( \int \frac{dx}{\sqrt{1+x} - \sqrt{x}} \).
Answer: To integrate a function with square roots in the denominator, it is common to rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator.
\( \int \frac{dx}{\sqrt{1+x} - \sqrt{x}} = \int \frac{1}{\sqrt{1+x} - \sqrt{x}} \cdot \frac{\sqrt{1+x} + \sqrt{x}}{\sqrt{1+x} + \sqrt{x}} dx \)
Using the difference of squares formula \( (a-b)(a+b) = a^2-b^2 \) in the denominator:
\( = \int \frac{\sqrt{1+x} + \sqrt{x}}{(\sqrt{1+x})^2 - (\sqrt{x})^2} dx \)
\( = \int \frac{\sqrt{1+x} + \sqrt{x}}{(1+x) - x} dx \)
\( = \int \frac{\sqrt{1+x} + \sqrt{x}}{1} dx \)
\( = \int ((1+x)^{1/2} + x^{1/2}) dx \)
Now, integrate each term using the power rule.
\( = \frac{(1+x)^{1/2+1}}{1/2+1} + \frac{x^{1/2+1}}{1/2+1} + C \)
\( = \frac{(1+x)^{3/2}}{3/2} + \frac{x^{3/2}}{3/2} + C \)
\( = \frac{2}{3}(1+x)^{3/2} + \frac{2}{3}x^{3/2} + C \)
In simple words: To get rid of the roots at the bottom, multiply the top and bottom by the "opposite" of the bottom part. This will simplify the bottom to just 1. Then, you can easily integrate the remaining terms on top.

🎯 Exam Tip: Always rationalize the denominator when dealing with integrals involving sums or differences of square roots. This converts the integrand into a form suitable for the power rule of integration.

Question 15. Integrate \( \int (\tan^2 x - \cot^2 x) dx \).
Answer: We use the trigonometric identities \( \tan^2 x = \sec^2 x - 1 \) and \( \cot^2 x = \operatorname{cosec}^2 x - 1 \) to transform the integrand into standard integrable forms. This helps make the terms manageable.
\( \int (\tan^2 x - \cot^2 x) dx = \int ((\sec^2 x - 1) - (\operatorname{cosec}^2 x - 1)) dx \)
\( = \int (\sec^2 x - 1 - \operatorname{cosec}^2 x + 1) dx \)
\( = \int (\sec^2 x - \operatorname{cosec}^2 x) dx \)
Now, integrate each term separately.
\( = \int \sec^2 x dx - \int \operatorname{cosec}^2 x dx \)
\( = \tan x - (-\cot x) + C \)
\( = \tan x + \cot x + C \)
In simple words: Replace \( \tan^2 x \) and \( \cot^2 x \) with their standard math identities involving \( \sec^2 x \) and \( \operatorname{cosec}^2 x \). The 'minus 1' parts will cancel out. Then, integrate the remaining secant and cosecant squared terms.

🎯 Exam Tip: When integrating differences of squared trigonometric functions, substitute their respective identities involving \( \sec^2 x \) and \( \operatorname{cosec}^2 x \) early on. This simplification usually leads to a straightforward integration.

Question 16. Integrate \( \int \frac{\sin x}{1+\sin x} dx \).
Answer: To integrate this function, we can multiply the numerator and denominator by the conjugate of the denominator, which is \( (1-\sin x) \). This helps convert the denominator into \( \cos^2 x \).
\( \int \frac{\sin x}{1+\sin x} dx = \int \frac{\sin x}{1+\sin x} \cdot \frac{1-\sin x}{1-\sin x} dx \)
\( = \int \frac{\sin x - \sin^2 x}{1-\sin^2 x} dx \)
Use the identity \( 1-\sin^2 x = \cos^2 x \).
\( = \int \frac{\sin x - \sin^2 x}{\cos^2 x} dx \)
Split the fraction into two separate terms.
\( = \int \left(\frac{\sin x}{\cos^2 x} - \frac{\sin^2 x}{\cos^2 x}\right) dx \)
Rewrite in terms of \( \sec x \), \( \tan x \), and \( \tan^2 x \).
\( = \int \left(\frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} - \tan^2 x\right) dx \)
\( = \int (\sec x \tan x - \tan^2 x) dx \)
Now, use the identity \( \tan^2 x = \sec^2 x - 1 \).
\( = \int (\sec x \tan x - (\sec^2 x - 1)) dx \)
\( = \int (\sec x \tan x - \sec^2 x + 1) dx \)
Integrate each term.
\( = \int \sec x \tan x dx - \int \sec^2 x dx + \int 1 dx \)
\( = \sec x - \tan x + x + C \)
In simple words: Multiply the top and bottom by \( (1-\sin x) \). Simplify the bottom using a math rule. Then, split the fraction into parts, change to secant and tangent terms, and integrate them using standard rules.

🎯 Exam Tip: Rationalizing the denominator with \( (1-\sin x) \) is a common technique for integrals involving \( (1+\sin x) \) or \( (1+\cos x) \). Also, remember to handle \( \tan^2 x \) by converting it to \( \sec^2 x - 1 \).

Question 17. Integrate \( \int \frac{1}{1-\cos x} dx \).
Answer: To integrate this, we can multiply the numerator and denominator by the conjugate of the denominator, which is \( (1+\cos x) \). This transforms the denominator into \( \sin^2 x \).
\( \int \frac{1}{1-\cos x} dx = \int \frac{1}{1-\cos x} \cdot \frac{1+\cos x}{1+\cos x} dx \)
\( = \int \frac{1+\cos x}{1-\cos^2 x} dx \)
Use the identity \( 1-\cos^2 x = \sin^2 x \).
\( = \int \frac{1+\cos x}{\sin^2 x} dx \)
Split the fraction into two separate terms.
\( = \int \left(\frac{1}{\sin^2 x} + \frac{\cos x}{\sin^2 x}\right) dx \)
Rewrite in terms of \( \operatorname{cosec}^2 x \) and \( \cot x \operatorname{cosec} x \).
\( = \int (\operatorname{cosec}^2 x + \cot x \operatorname{cosec} x) dx \)
Integrate each term separately.
\( = \int \operatorname{cosec}^2 x dx + \int \cot x \operatorname{cosec} x dx \)
\( = -\cot x - \operatorname{cosec} x + C \)
In simple words: Multiply the top and bottom by \( (1+\cos x) \). This makes the bottom \( \sin^2 x \). Then, break the fraction into two parts using standard trig terms like cosecant and cotangent, and integrate each one.

🎯 Exam Tip: When faced with \( \frac{1}{1 \pm \cos x} \) or \( \frac{1}{1 \pm \sin x} \), always multiply by the conjugate to transform the denominator into \( \sin^2 x \) or \( \cos^2 x \), which simplifies the integrand using basic trigonometric identities.

Question 18. Integrate \( \int \left[1 + \frac{1}{1+x^2} + \frac{3}{x\sqrt{x^2-1}} + 2^x\right] dx \).
Answer: We can integrate each term in the sum separately, as each term corresponds to a standard integral formula. This is possible because integration is linear.
\( \int \left[1 + \frac{1}{1+x^2} + \frac{3}{x\sqrt{x^2-1}} + 2^x\right] dx \)
\( = \int 1 dx + \int \frac{1}{1+x^2} dx + \int \frac{3}{x\sqrt{x^2-1}} dx + \int 2^x dx \)
\( = \int 1 dx + \int \frac{1}{1+x^2} dx + 3 \int \frac{1}{x\sqrt{x^2-1}} dx + \int 2^x dx \)
Now, apply the respective integral formulas:
\( \int 1 dx = x \)
\( \int \frac{1}{1+x^2} dx = \tan^{-1} x \)
\( \int \frac{1}{x\sqrt{x^2-1}} dx = \sec^{-1} x \)
\( \int 2^x dx = \frac{2^x}{\log_e 2} \)
Combining these, we get:
\( = x + \tan^{-1} x + 3 \sec^{-1} x + \frac{2^x}{\log_e 2} + C \)
In simple words: This problem asks us to integrate four different types of terms added together. We simply integrate each one using its own special rule: the integral of 1 is x, for \( 1/(1+x^2) \) it's inverse tangent, for \( 1/(x\sqrt{x^2-1}) \) it's inverse secant, and for \( 2^x \) it's \( 2^x \) divided by \( \log_e 2 \).

🎯 Exam Tip: When integrating sums or differences of multiple functions, remember that you can integrate each term independently. It's crucial to recognize and apply the correct standard integral formula for each part.

Question 19. Integrate \( \int \cot x (\tan x - \operatorname{cosec} x) dx \).
Answer: First, distribute \( \cot x \) to each term inside the parenthesis. This helps simplify the expression using trigonometric identities.
\( \int \cot x (\tan x - \operatorname{cosec} x) dx = \int (\cot x \tan x - \cot x \operatorname{cosec} x) dx \)
Use the identity \( \cot x \tan x = 1 \).
\( = \int (1 - \cot x \operatorname{cosec} x) dx \)
Now, integrate each term separately.
\( = \int 1 dx - \int \cot x \operatorname{cosec} x dx \)
\( = x - (-\operatorname{cosec} x) + C \)
\( = x + \operatorname{cosec} x + C \)
In simple words: Multiply \( \cot x \) by \( \tan x \) and then by \( -\operatorname{cosec} x \). Since \( \cot x \tan x \) equals 1, you're left with \( 1 - \cot x \operatorname{cosec} x \). Integrate these two simple terms separately.

🎯 Exam Tip: Always expand products of trigonometric functions. This often reveals fundamental identities (like \( \cot x \tan x = 1 \)) that simplify the integrand into easily integrable forms, such as \( 1 \) or \( \cot x \operatorname{cosec} x \).

Question 20. Integrate \( \int \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 dx \).
Answer: First, expand the squared term using the formula \( (a+b)^2 = a^2+2ab+b^2 \). This will simplify the expression into powers of x, which are easy to integrate.
\( \int \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 dx = \int \left((\sqrt{x})^2 + 2\sqrt{x}\frac{1}{\sqrt{x}} + \left(\frac{1}{\sqrt{x}}\right)^2\right) dx \)
\( = \int \left(x + 2 + \frac{1}{x}\right) dx \)
Now, integrate each term separately using the power rule for x and the special rule for \( 1/x \).
\( = \int x dx + \int 2 dx + \int \frac{1}{x} dx \)
\( = \frac{x^2}{2} + 2x + \log|x| + C \)
In simple words: First, open up the square bracket. The terms \( \sqrt{x} \) and \( 1/\sqrt{x} \) will simplify to x and 1/x, with a middle term of 2. Then, integrate each simple term: \( x \), \( 2 \), and \( 1/x \).

🎯 Exam Tip: Always expand squared binomials or products before integrating. This often breaks down complex expressions into simpler power functions and constant terms that are easily integrable.

Question 21. Integrate \( \int \log_x x dx \).
Answer: We use the fundamental property of logarithms that \( \log_b b = 1 \). Therefore, \( \log_x x \) simplifies to 1 for \( x > 0 \) and \( x \neq 1 \). This reduces the integral to a very simple form.
\( \int \log_x x dx = \int 1 dx \)
Now, integrate the constant 1 with respect to x.
\( = x + C \)
In simple words: The number \( \log_x x \) is always equal to 1, as long as x is not 1 and is positive. So the problem just becomes integrating the number 1, which gives x.

🎯 Exam Tip: Recognize and apply basic logarithmic properties (like \( \log_b b = 1 \)) to simplify integrands. This can transform seemingly complex integrals into elementary ones.

Question 22. Integrate \( \int \sqrt{1 + \cos 2x} dx \).
Answer: We use the double angle identity for cosine, \( 1 + \cos 2x = 2 \cos^2 x \). This allows us to simplify the expression under the square root and make it integrable.
\( \int \sqrt{1 + \cos 2x} dx = \int \sqrt{2 \cos^2 x} dx \)
\( = \int \sqrt{2} |\cos x| dx \)
For typical integration problems of this nature (unless a specific interval is given), we often assume \( \cos x \ge 0 \) within the integration range to simplify \( |\cos x| \) to \( \cos x \).
\( = \sqrt{2} \int \cos x dx \)
Now, integrate \( \cos x \).
\( = \sqrt{2} \sin x + C \)
In simple words: Use the math rule \( 1 + \cos 2x = 2 \cos^2 x \). This makes the square root easier to solve, leaving \( \sqrt{2} \cos x \). Then, integrate \( \cos x \) to get \( \sin x \).

🎯 Exam Tip: When \( (1+\cos 2x) \) or \( (1-\cos 2x) \) appear under a square root, use the identities \( 1+\cos 2x = 2\cos^2 x \) or \( 1-\cos 2x = 2\sin^2 x \) to eliminate the square root and simplify the integrand.

Question 23. Integrate \( \int \frac{\cos 2x}{\sin^2 x \cos^2 x} dx \).
Answer: We use the double angle identity \( \cos 2x = \cos^2 x - \sin^2 x \) in the numerator. This allows us to split the fraction into two simpler, integrable terms.
\( \int \frac{\cos 2x}{\sin^2 x \cos^2 x} dx = \int \frac{\cos^2 x - \sin^2 x}{\sin^2 x \cos^2 x} dx \)
Split the fraction into two parts.
\( = \int \left(\frac{\cos^2 x}{\sin^2 x \cos^2 x} - \frac{\sin^2 x}{\sin^2 x \cos^2 x}\right) dx \)
Simplify each part.
\( = \int \left(\frac{1}{\sin^2 x} - \frac{1}{\cos^2 x}\right) dx \)
Rewrite in terms of \( \operatorname{cosec}^2 x \) and \( \sec^2 x \).
\( = \int (\operatorname{cosec}^2 x - \sec^2 x) dx \)
Now, integrate each term separately.
\( = \int \operatorname{cosec}^2 x dx - \int \sec^2 x dx \)
\( = -\cot x - \tan x + C \)
In simple words: Change \( \cos 2x \) to \( \cos^2 x - \sin^2 x \). Then, break the fraction into two smaller parts and simplify each. Finally, integrate the cosecant squared and secant squared terms.

🎯 Exam Tip: When \( \cos 2x \) is in the numerator of a fraction with \( \sin^2 x \cos^2 x \) in the denominator, use the identity \( \cos 2x = \cos^2 x - \sin^2 x \) to split the fraction. This leads to common integrable forms involving \( \operatorname{cosec}^2 x \) and \( \sec^2 x \).

Question 24. Integrate \( \int \frac{3 \cos x + 4}{\sin^2 x} dx \).
Answer: We can split the fraction into two separate terms by dividing each term in the numerator by the common denominator \( \sin^2 x \). This transforms the integrand into standard trigonometric forms.
\( \int \frac{3 \cos x + 4}{\sin^2 x} dx = \int \left(\frac{3 \cos x}{\sin^2 x} + \frac{4}{\sin^2 x}\right) dx \)
Rewrite the terms using \( \cot x \), \( \operatorname{cosec} x \), and \( \operatorname{cosec}^2 x \).
\( = \int \left(3 \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} + 4 \cdot \frac{1}{\sin^2 x}\right) dx \)
\( = \int (3 \cot x \operatorname{cosec} x + 4 \operatorname{cosec}^2 x) dx \)
Now, integrate each term separately.
\( = 3 \int \cot x \operatorname{cosec} x dx + 4 \int \operatorname{cosec}^2 x dx \)
\( = 3 (-\operatorname{cosec} x) + 4 (-\cot x) + C \)
\( = -3 \operatorname{cosec} x - 4 \cot x + C \)
In simple words: Break the fraction into two separate parts. For the first part, change it to \( \cot x \operatorname{cosec} x \). For the second part, change it to \( \operatorname{cosec}^2 x \). Then, integrate each of these parts using their specific rules.

🎯 Exam Tip: When a fraction has a sum in the numerator and a single term in the denominator, always split it into separate fractions. This often simplifies the terms into standard integrable trigonometric functions like \( \cot x \operatorname{cosec} x \) and \( \operatorname{cosec}^2 x \).