RBSE Solutions Class 12 Maths Chapter 9 समाकलन Exercise 9.6

निम्न फलनों का x के सापेक्ष समाकलन कीजिए

Question 1. (a) ∫x cos x dx (b) ∫x sec²x dx
Answer:
(a) For \( \int x \cos x \, dx \), we use integration by parts, taking \( x \) as the first function and \( \cos x \) as the second function.
\( \int x \cos x \, dx = x \int \cos x \, dx - \int \left( \frac{d}{dx}(x) \int \cos x \, dx \right) \, dx \)
\( = x (\sin x) - \int (1)(\sin x) \, dx \)
\( = x \sin x - \int \sin x \, dx \)
\( = x \sin x - (-\cos x) + C \)
\( = x \sin x + \cos x + C \)
(b) For \( \int x \sec^2 x \, dx \), we again use integration by parts, taking \( x \) as the first function and \( \sec^2 x \) as the second function.
\( \int x \sec^2 x \, dx = x \int \sec^2 x \, dx - \int \left( \frac{d}{dx}(x) \int \sec^2 x \, dx \right) \, dx \)
\( = x (\tan x) - \int (1)(\tan x) \, dx \)
\( = x \tan x - \int \tan x \, dx \)
\( = x \tan x - (-\log |\cos x|) + C \)
\( = x \tan x + \log |\cos x| + C \)
This can also be written as: \( x \tan x - \log |\sec x| + C \).
In simple words: To solve these, we use a special method called integration by parts. It helps break down complex multiplication in integrals into easier steps. The formula is: Integral of (first * second) = first * Integral of second - Integral of (derivative of first * Integral of second). We apply this rule step-by-step.

🎯 Exam Tip: Remember to choose the "first" and "second" functions wisely in integration by parts using the "ILATE" rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential) to simplify calculations.

Question 2. (a) ∫x³e⁻ˣ dx (b) ∫x³ sinx dx
Answer:
(a) For \( \int x^3 e^{-x} \, dx \), we use integration by parts multiple times. Let \( u = x^3 \) and \( dv = e^{-x} \, dx \). Then \( du = 3x^2 \, dx \) and \( v = -e^{-x} \).
\( \int x^3 e^{-x} \, dx = x^3 (-e^{-x}) - \int (3x^2) (-e^{-x}) \, dx \)
\( = -x^3 e^{-x} + 3 \int x^2 e^{-x} \, dx \quad \ldots (i) \)
Now, let's solve \( \int x^2 e^{-x} \, dx \). Let \( u = x^2 \) and \( dv = e^{-x} \, dx \). Then \( du = 2x \, dx \) and \( v = -e^{-x} \).
\( \int x^2 e^{-x} \, dx = x^2 (-e^{-x}) - \int (2x) (-e^{-x}) \, dx \)
\( = -x^2 e^{-x} + 2 \int x e^{-x} \, dx \quad \ldots (ii) \)
Next, let's solve \( \int x e^{-x} \, dx \). Let \( u = x \) and \( dv = e^{-x} \, dx \). Then \( du = 1 \, dx \) and \( v = -e^{-x} \).
\( \int x e^{-x} \, dx = x (-e^{-x}) - \int (1) (-e^{-x}) \, dx \)
\( = -x e^{-x} + \int e^{-x} \, dx \)
\( = -x e^{-x} - e^{-x} + C_1 \quad \ldots (iii) \)
Now, substitute (iii) into (ii):
\( \int x^2 e^{-x} \, dx = -x^2 e^{-x} + 2(-x e^{-x} - e^{-x} + C_1) \)
\( = -x^2 e^{-x} - 2x e^{-x} - 2e^{-x} + 2C_1 \)
Finally, substitute this back into (i):
\( \int x^3 e^{-x} \, dx = -x^3 e^{-x} + 3(-x^2 e^{-x} - 2x e^{-x} - 2e^{-x} + 2C_1) \)
\( = -x^3 e^{-x} - 3x^2 e^{-x} - 6x e^{-x} - 6e^{-x} + 6C_1 \)
\( = -e^{-x} (x^3 + 3x^2 + 6x + 6) + C \)
(b) For \( \int x^3 \sin x \, dx \), we use integration by parts repeatedly. Let \( u = x^3 \) and \( dv = \sin x \, dx \). Then \( du = 3x^2 \, dx \) and \( v = -\cos x \).
\( \int x^3 \sin x \, dx = x^3 (-\cos x) - \int (3x^2) (-\cos x) \, dx \)
\( = -x^3 \cos x + 3 \int x^2 \cos x \, dx \)
Now, for \( \int x^2 \cos x \, dx \), let \( u = x^2 \) and \( dv = \cos x \, dx \). Then \( du = 2x \, dx \) and \( v = \sin x \).
\( \int x^2 \cos x \, dx = x^2 (\sin x) - \int (2x) (\sin x) \, dx \)
\( = x^2 \sin x - 2 \int x \sin x \, dx \)
Next, for \( \int x \sin x \, dx \), let \( u = x \) and \( dv = \sin x \, dx \). Then \( du = 1 \, dx \) and \( v = -\cos x \).
\( \int x \sin x \, dx = x (-\cos x) - \int (1) (-\cos x) \, dx \)
\( = -x \cos x + \int \cos x \, dx \)
\( = -x \cos x + \sin x \)
Substitute back:
\( \int x^3 \sin x \, dx = -x^3 \cos x + 3(x^2 \sin x - 2(-x \cos x + \sin x)) + C \)
\( = -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x + C \)
In simple words: These problems need integration by parts many times, which can be a bit long. You take one part of the multiplication and differentiate it, and the other part and integrate it, then combine them. You keep doing this until the integral is easy to solve.

🎯 Exam Tip: When integrating \( x^n e^{ax} \) or \( x^n \sin(ax) \), you often need to apply integration by parts \( n \) times. Keep track of signs and constants carefully.

Question 3. (a) ∫x³ (log x)² dx (b) ∫x³ex² dx.
Answer:
(a) For \( \int x^3 (\log x)^2 \, dx \), we use integration by parts, taking \( (\log x)^2 \) as the first function and \( x^3 \) as the second.
\( \int (\log x)^2 x^3 \, dx = (\log x)^2 \int x^3 \, dx - \int \left( \frac{d}{dx}((\log x)^2) \int x^3 \, dx \right) \, dx \)
\( = (\log x)^2 \left( \frac{x^4}{4} \right) - \int \left( 2 \log x \cdot \frac{1}{x} \right) \left( \frac{x^4}{4} \right) \, dx \)
\( = \frac{x^4}{4} (\log x)^2 - \frac{1}{2} \int x^3 \log x \, dx \)
Now, we apply integration by parts to \( \int x^3 \log x \, dx \), taking \( \log x \) as the first function and \( x^3 \) as the second.
\( \int x^3 \log x \, dx = \log x \int x^3 \, dx - \int \left( \frac{d}{dx}(\log x) \int x^3 \, dx \right) \, dx \)
\( = \log x \left( \frac{x^4}{4} \right) - \int \left( \frac{1}{x} \right) \left( \frac{x^4}{4} \right) \, dx \)
\( = \frac{x^4}{4} \log x - \frac{1}{4} \int x^3 \, dx \)
\( = \frac{x^4}{4} \log x - \frac{1}{4} \left( \frac{x^4}{4} \right) = \frac{x^4}{4} \log x - \frac{x^4}{16} \)
Substitute this back into the main expression:
\( \int x^3 (\log x)^2 \, dx = \frac{x^4}{4} (\log x)^2 - \frac{1}{2} \left( \frac{x^4}{4} \log x - \frac{x^4}{16} \right) + C \)
\( = \frac{x^4}{4} (\log x)^2 - \frac{x^4}{8} \log x + \frac{x^4}{32} + C \)
\( = \frac{x^4}{4} \left( (\log x)^2 - \frac{1}{2} \log x + \frac{1}{8} \right) + C \)
(b) For \( \int x^3 e^{x^2} \, dx \), we use substitution and then integration by parts.
Let \( t = x^2 \). Then \( \frac{dt}{dx} = 2x \implies dt = 2x \, dx \implies x \, dx = \frac{dt}{2} \).
Rewrite the integral as \( \int x^2 e^{x^2} x \, dx \).
Substitute \( t \) and \( dt \):
\( \int t e^t \frac{dt}{2} = \frac{1}{2} \int t e^t \, dt \)
Now, apply integration by parts to \( \int t e^t \, dt \). Let \( u = t \) and \( dv = e^t \, dt \). Then \( du = 1 \, dt \) and \( v = e^t \).
\( \int t e^t \, dt = t e^t - \int (1) e^t \, dt \)
\( = t e^t - e^t = e^t (t-1) \)
Substitute this back:
\( \frac{1}{2} \int t e^t \, dt = \frac{1}{2} e^t (t-1) + C \)
Finally, substitute back \( t = x^2 \):
\( = \frac{1}{2} e^{x^2} (x^2-1) + C \)
In simple words: For part (a), we use the integration by parts rule two times. For part (b), we first change the variable to make it simpler, and then use integration by parts once. Both methods help us find the integral step-by-step.

🎯 Exam Tip: When dealing with integrals involving \( (\log x)^n \), always consider \( (\log x)^n \) as the first function in integration by parts. For integrals like \( x^3 e^{x^2} \), look for suitable substitutions that simplify the exponent.

Question 4. (a) ∫ e²ˣ eᵉˣ dx (b) ∫ (log x)² dx
Answer:
(a) For \( \int e^{2x} e^{e^x} \, dx \), we can rewrite \( e^{2x} \) as \( e^x \cdot e^x \) and use substitution.
Let \( t = e^x \). Then \( \frac{dt}{dx} = e^x \implies dt = e^x \, dx \).
The integral becomes \( \int e^x \cdot e^x \cdot e^{e^x} \, dx = \int (e^x) e^{e^x} (e^x \, dx) \).
Substitute \( t \) and \( dt \):
\( \int t e^t \, dt \)
Now, apply integration by parts to \( \int t e^t \, dt \). Let \( u = t \) and \( dv = e^t \, dt \). Then \( du = 1 \, dt \) and \( v = e^t \).
\( \int t e^t \, dt = t e^t - \int (1) e^t \, dt \)
\( = t e^t - e^t + C \)
\( = e^t (t-1) + C \)
Finally, substitute back \( t = e^x \):
\( = e^{e^x} (e^x-1) + C \)
(b) For \( \int (\log x)^2 \, dx \), we treat it as \( \int (\log x)^2 \cdot 1 \, dx \) and use integration by parts. Take \( (\log x)^2 \) as the first function and \( 1 \) as the second.
\( \int (\log x)^2 \cdot 1 \, dx = (\log x)^2 \int 1 \, dx - \int \left( \frac{d}{dx}((\log x)^2) \int 1 \, dx \right) \, dx \)
\( = (\log x)^2 (x) - \int \left( 2 \log x \cdot \frac{1}{x} \right) (x) \, dx \)
\( = x (\log x)^2 - 2 \int \log x \, dx \)
Now, we apply integration by parts to \( \int \log x \, dx \), treating it as \( \int \log x \cdot 1 \, dx \). Take \( \log x \) as the first function and \( 1 \) as the second.
\( \int \log x \cdot 1 \, dx = \log x \int 1 \, dx - \int \left( \frac{d}{dx}(\log x) \int 1 \, dx \right) \, dx \)
\( = \log x (x) - \int \left( \frac{1}{x} \right) (x) \, dx \)
\( = x \log x - \int 1 \, dx \)
\( = x \log x - x \)
Substitute this back into the main expression:
\( \int (\log x)^2 \, dx = x (\log x)^2 - 2(x \log x - x) + C \)
\( = x (\log x)^2 - 2x \log x + 2x + C \)
In simple words: Part (a) needs a substitution first to simplify the problem, then uses integration by parts. Part (b) directly uses integration by parts twice to solve the integral of a squared logarithm. Both methods are standard for these types of questions.

🎯 Exam Tip: When you see \( \int f(e^x) e^x dx \), consider \( t = e^x \) as a substitution. Also, remember that \( \int \log x \, dx = x \log x - x + C \) is a common result that you might need to use or derive.

Question 5. (a) ∫cos¯¹x dx (b) ∫cosec¯¹ \( \sqrt{\frac{x+a}{x}} \) dx
Answer:
(a) For \( \int \cos^{-1}x \, dx \), we treat it as \( \int \cos^{-1}x \cdot 1 \, dx \) and use integration by parts. Let \( u = \cos^{-1}x \) (first function) and \( dv = 1 \, dx \) (second function). Then \( du = \frac{-1}{\sqrt{1-x^2}} \, dx \) and \( v = x \).
\( \int \cos^{-1}x \, dx = x \cos^{-1}x - \int (x) \left( \frac{-1}{\sqrt{1-x^2}} \right) \, dx \)
\( = x \cos^{-1}x + \int \frac{x}{\sqrt{1-x^2}} \, dx \)
To solve \( \int \frac{x}{\sqrt{1-x^2}} \, dx \), let \( t = 1-x^2 \). Then \( dt = -2x \, dx \implies x \, dx = -\frac{dt}{2} \).
\( \int \frac{x}{\sqrt{1-x^2}} \, dx = \int \frac{-dt/2}{\sqrt{t}} = -\frac{1}{2} \int t^{-1/2} \, dt \)
\( = -\frac{1}{2} \left( \frac{t^{1/2}}{1/2} \right) = -t^{1/2} = -\sqrt{t} \)
Substitute back \( t = 1-x^2 \): \( -\sqrt{1-x^2} \).
So, \( \int \cos^{-1}x \, dx = x \cos^{-1}x - \sqrt{1-x^2} + C \).
(b) For \( \int \operatorname{cosec}^{-1} \sqrt{\frac{x+a}{x}} \, dx \), we first convert \( \operatorname{cosec}^{-1} \) to \( \sin^{-1} \).
\( \operatorname{cosec}^{-1} y = \sin^{-1} \left( \frac{1}{y} \right) \).
So, \( \operatorname{cosec}^{-1} \sqrt{\frac{x+a}{x}} = \sin^{-1} \left( \frac{1}{\sqrt{\frac{x+a}{x}}} \right) = \sin^{-1} \left( \sqrt{\frac{x}{x+a}} \right) \).
The integral becomes \( \int \sin^{-1} \left( \sqrt{\frac{x}{x+a}} \right) \, dx \).
This problem is quite involved and requires a trigonometric substitution for \( x \). Let \( x = a \tan^2 \theta \). Then \( dx = 2a \tan \theta \sec^2 \theta \, d\theta \).
\( \sqrt{\frac{x}{x+a}} = \sqrt{\frac{a \tan^2 \theta}{a \tan^2 \theta + a}} = \sqrt{\frac{\tan^2 \theta}{\tan^2 \theta + 1}} = \sqrt{\frac{\tan^2 \theta}{\sec^2 \theta}} = \sqrt{\sin^2 \theta} = \sin \theta \).
So, \( \sin^{-1} \left( \sqrt{\frac{x}{x+a}} \right) = \sin^{-1}(\sin \theta) = \theta \).
The integral becomes \( \int \theta (2a \tan \theta \sec^2 \theta) \, d\theta = 2a \int \theta \tan \theta \sec^2 \theta \, d\theta \).
This is an integral that needs integration by parts. Let \( u = \theta \) and \( dv = \tan \theta \sec^2 \theta \, d\theta \).
\( \int \tan \theta \sec^2 \theta \, d\theta \). Let \( w = \tan \theta \), then \( dw = \sec^2 \theta \, d\theta \). So, \( \int w \, dw = \frac{w^2}{2} = \frac{\tan^2 \theta}{2} \).
Thus, \( v = \frac{\tan^2 \theta}{2} \).
\( 2a \left[ \theta \frac{\tan^2 \theta}{2} - \int (1) \frac{\tan^2 \theta}{2} \, d\theta \right] \)
\( = a \left[ \theta \tan^2 \theta - \int (\sec^2 \theta - 1) \, d\theta \right] \)
\( = a \left[ \theta \tan^2 \theta - (\tan \theta - \theta) \right] + C \)
\( = a \left[ \theta \tan^2 \theta - \tan \theta + \theta \right] + C \)
Substitute back \( \theta = \tan^{-1} \sqrt{x/a} \) and \( \tan^2 \theta = x/a \).
\( = a \left[ \tan^{-1} \sqrt{x/a} \cdot \frac{x}{a} - \sqrt{x/a} + \tan^{-1} \sqrt{x/a} \right] + C \)
\( = x \tan^{-1} \sqrt{x/a} - a \sqrt{x/a} + a \tan^{-1} \sqrt{x/a} + C \)
\( = (x+a) \tan^{-1} \sqrt{x/a} - \sqrt{ax} + C \)
In simple words: For part (a), we use integration by parts for the inverse cosine function. For part (b), we first change the inverse cosec to inverse sin, then use a special substitution (like \( x = a \tan^2 \theta \)) to simplify the expression before applying integration by parts. These methods help us solve integrals that look complex at first glance.

🎯 Exam Tip: For inverse trigonometric functions like \( \int \cos^{-1}x \, dx \), always multiply by 1 and use integration by parts. For expressions like \( \sqrt{\frac{x}{x+a}} \), look for substitutions like \( x=a\tan^2\theta \) or \( x=a\cot^2\theta \) to simplify the square root.

Question 6. (a) ∫ sin¯¹ (3x - 4x³) dx (b) ∫ x/(1+cos x) dx
Answer:
(a) For \( \int \sin^{-1} (3x - 4x^3) \, dx \), we recognize the identity \( \sin(3\theta) = 3\sin\theta - 4\sin^3\theta \).
Let \( x = \sin\theta \). Then \( dx = \cos\theta \, d\theta \).
\( 3x - 4x^3 = 3\sin\theta - 4\sin^3\theta = \sin(3\theta) \).
So, \( \sin^{-1}(3x - 4x^3) = \sin^{-1}(\sin(3\theta)) = 3\theta \).
The integral becomes \( \int 3\theta \cos\theta \, d\theta = 3 \int \theta \cos\theta \, d\theta \).
Now, apply integration by parts to \( \int \theta \cos\theta \, d\theta \). Let \( u = \theta \) (first function) and \( dv = \cos\theta \, d\theta \) (second function). Then \( du = 1 \, d\theta \) and \( v = \sin\theta \).
\( \int \theta \cos\theta \, d\theta = \theta \sin\theta - \int (1) \sin\theta \, d\theta \)
\( = \theta \sin\theta - (-\cos\theta) \)
\( = \theta \sin\theta + \cos\theta \)
Substitute this back into the main expression:
\( 3 \int \theta \cos\theta \, d\theta = 3(\theta \sin\theta + \cos\theta) + C \)
Substitute back \( \theta = \sin^{-1}x \). Since \( x = \sin\theta \), \( \cos\theta = \sqrt{1-\sin^2\theta} = \sqrt{1-x^2} \).
\( = 3(x \sin^{-1}x + \sqrt{1-x^2}) + C \)
(b) For \( \int \frac{x}{1+\cos x} \, dx \), we use the half-angle identity \( 1+\cos x = 2\cos^2(x/2) \).
\( \int \frac{x}{1+\cos x} \, dx = \int \frac{x}{2\cos^2(x/2)} \, dx = \frac{1}{2} \int x \sec^2(x/2) \, dx \)
Now, apply integration by parts. Let \( u = x \) (first function) and \( dv = \sec^2(x/2) \, dx \) (second function).
Then \( du = 1 \, dx \) and \( v = \int \sec^2(x/2) \, dx = 2\tan(x/2) \).
\( \frac{1}{2} \left[ x (2\tan(x/2)) - \int (1) (2\tan(x/2)) \, dx \right] \)
\( = \frac{1}{2} \left[ 2x \tan(x/2) - 2 \int \tan(x/2) \, dx \right] \)
\( = x \tan(x/2) - \int \tan(x/2) \, dx \)
The integral of \( \tan(ax) \) is \( \frac{1}{a} \log|\sec(ax)| \). So, \( \int \tan(x/2) \, dx = \frac{1}{1/2} \log|\sec(x/2)| = 2 \log|\sec(x/2)| \).
Therefore, \( = x \tan(x/2) - 2 \log|\sec(x/2)| + C \).
In simple words: Part (a) uses a trigonometric identity to simplify the inverse sine term, then solves it with integration by parts. Part (b) uses another trigonometric identity to simplify the denominator, making it easier to integrate using integration by parts. Knowing these identities is key to solving such problems.

🎯 Exam Tip: Always look for trigonometric identities like \( \sin(3\theta) \) or \( 1+\cos x \) to simplify complex inverse trigonometric or rational function integrals before applying integration by parts.

Question 7. (a) ∫tan¯¹ \( \sqrt{\frac{1-x}{1+x}} \) dx (b) ∫ cos √x dx
Answer:
(a) For \( \int \tan^{-1} \sqrt{\frac{1-x}{1+x}} \, dx \), we use a trigonometric substitution to simplify the square root.
Let \( x = \cos\theta \). Then \( dx = -\sin\theta \, d\theta \).
\( \sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos\theta}{1+\cos\theta}} = \sqrt{\frac{2\sin^2(\theta/2)}{2\cos^2(\theta/2)}} = \sqrt{\tan^2(\theta/2)} = \tan(\theta/2) \).
So, \( \tan^{-1} \sqrt{\frac{1-x}{1+x}} = \tan^{-1}(\tan(\theta/2)) = \theta/2 \).
The integral becomes \( \int \frac{\theta}{2} (-\sin\theta) \, d\theta = -\frac{1}{2} \int \theta \sin\theta \, d\theta \).
Now, apply integration by parts to \( \int \theta \sin\theta \, d\theta \). Let \( u = \theta \) (first function) and \( dv = \sin\theta \, d\theta \) (second function). Then \( du = 1 \, d\theta \) and \( v = -\cos\theta \).
\( \int \theta \sin\theta \, d\theta = \theta (-\cos\theta) - \int (1) (-\cos\theta) \, d\theta \)
\( = -\theta \cos\theta + \int \cos\theta \, d\theta \)
\( = -\theta \cos\theta + \sin\theta \)
Substitute this back into the main expression:
\( -\frac{1}{2} (-\theta \cos\theta + \sin\theta) + C = \frac{1}{2} (\theta \cos\theta - \sin\theta) + C \).
Substitute back \( \theta = \cos^{-1}x \). Since \( x = \cos\theta \), \( \sin\theta = \sqrt{1-\cos^2\theta} = \sqrt{1-x^2} \).
\( = \frac{1}{2} (x \cos^{-1}x - \sqrt{1-x^2}) + C \).
(b) For \( \int \cos \sqrt{x} \, dx \), we use substitution first.
Let \( t = \sqrt{x} \). Then \( t^2 = x \), so \( 2t \, dt = dx \).
The integral becomes \( \int \cos t (2t \, dt) = 2 \int t \cos t \, dt \).
Now, apply integration by parts to \( \int t \cos t \, dt \). Let \( u = t \) (first function) and \( dv = \cos t \, dt \) (second function). Then \( du = 1 \, dt \) and \( v = \sin t \).
\( \int t \cos t \, dt = t \sin t - \int (1) \sin t \, dt \)
\( = t \sin t - (-\cos t) \)
\( = t \sin t + \cos t \)
Substitute this back into the main 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 change \( x \) to \( \cos\theta \) to simplify the square root and the inverse tangent, then use integration by parts. For part (b), we replace \( \sqrt{x} \) with a new variable \( t \), which turns the integral into a form that can be solved using integration by parts. Both problems show how substitutions can make hard integrals easier.

🎯 Exam Tip: When you see \( \sqrt{\frac{1-x}{1+x}} \), try \( x=\cos\theta \) substitution. For \( \sqrt{x} \) inside a function, \( t=\sqrt{x} \) is a very effective substitution that turns it into an algebraic integral.

Question 8. (a) ∫ x/(1+sin x) dx (b) ∫ x² tan¯¹x dx
Answer:
(a) For \( \int \frac{x}{1+\sin x} \, dx \), we multiply the numerator and denominator by \( (1-\sin x) \) to rationalize.
\( \int \frac{x}{1+\sin x} \, dx = \int \frac{x(1-\sin x)}{(1+\sin x)(1-\sin x)} \, dx \)
\( = \int \frac{x(1-\sin x)}{1-\sin^2 x} \, dx = \int \frac{x(1-\sin x)}{\cos^2 x} \, dx \)
\( = \int \left( \frac{x}{\cos^2 x} - \frac{x \sin x}{\cos^2 x} \right) \, dx = \int (x \sec^2 x - x \tan x \sec x) \, dx \)
\( = \int x \sec^2 x \, dx - \int x \tan x \sec x \, dx \)
Let's solve \( I_1 = \int x \sec^2 x \, dx \) using integration by parts. Let \( u = x \) and \( dv = \sec^2 x \, dx \). Then \( du = 1 \, dx \) and \( v = \tan x \).
\( I_1 = x \tan x - \int (1) \tan x \, dx = x \tan x - (-\log |\cos x|) = x \tan x + \log |\cos x| \).
Let's solve \( I_2 = \int x \tan x \sec x \, dx \) using integration by parts. Let \( u = x \) and \( dv = \tan x \sec x \, dx \). Then \( du = 1 \, dx \) and \( v = \sec x \).
\( I_2 = x \sec x - \int (1) \sec x \, dx = x \sec x - \log |\sec x + \tan x| \).
Now, combine \( I_1 \) and \( I_2 \):
\( \int \frac{x}{1+\sin x} \, dx = I_1 - I_2 = (x \tan x + \log |\cos x|) - (x \sec x - \log |\sec x + \tan x|) + C \)
\( = x \tan x - x \sec x + \log |\cos x| + \log |\sec x + \tan x| + C \)
\( = x(\tan x - \sec x) + \log |\cos x (\sec x + \tan x)| + C \)
\( = x\left( \frac{\sin x - 1}{\cos x} \right) + \log |1 + \sin x| + C \)
(b) For \( \int x^2 \tan^{-1}x \, dx \), we use integration by parts, taking \( \tan^{-1}x \) as the first function and \( x^2 \) as the second.
\( \int x^2 \tan^{-1}x \, dx = \tan^{-1}x \int x^2 \, dx - \int \left( \frac{d}{dx}(\tan^{-1}x) \int x^2 \, dx \right) \, dx \)
\( = \tan^{-1}x \left( \frac{x^3}{3} \right) - \int \left( \frac{1}{1+x^2} \right) \left( \frac{x^3}{3} \right) \, dx \)
\( = \frac{x^3}{3} \tan^{-1}x - \frac{1}{3} \int \frac{x^3}{1+x^2} \, dx \)
To solve \( \int \frac{x^3}{1+x^2} \, dx \), we use polynomial division or rewrite the numerator.
\( \frac{x^3}{1+x^2} = \frac{x(x^2+1)-x}{1+x^2} = x - \frac{x}{1+x^2} \).
So, \( \int \frac{x^3}{1+x^2} \, dx = \int \left( x - \frac{x}{1+x^2} \right) \, dx = \int x \, dx - \int \frac{x}{1+x^2} \, dx \)
\( = \frac{x^2}{2} - \frac{1}{2} \int \frac{2x}{1+x^2} \, dx \)
The second integral is of the form \( \int \frac{f'(x)}{f(x)} \, dx = \log|f(x)| \). So, \( \int \frac{2x}{1+x^2} \, dx = \log|1+x^2| \).
Thus, \( \int \frac{x^3}{1+x^2} \, dx = \frac{x^2}{2} - \frac{1}{2} \log(1+x^2) \).
Substitute this back into the main expression:
\( \int x^2 \tan^{-1}x \, dx = \frac{x^3}{3} \tan^{-1}x - \frac{1}{3} \left( \frac{x^2}{2} - \frac{1}{2} \log(1+x^2) \right) + C \)
\( = \frac{x^3}{3} \tan^{-1}x - \frac{x^2}{6} + \frac{1}{6} \log(1+x^2) + C \)
In simple words: For part (a), we first make the denominator simpler by multiplying by its conjugate, which converts it to \( \cos^2 x \). Then we split the integral and use integration by parts twice. For part (b), we directly apply integration by parts and then use a trick to simplify the remaining fraction before integrating. Both problems show how different methods are used together.

🎯 Exam Tip: For rational functions of trigonometric terms, multiplying by the conjugate can often simplify the expression. For integrals involving \( \tan^{-1}x \) with an algebraic term, always take \( \tan^{-1}x \) as the first function in integration by parts. Remember that \( \frac{x^3}{1+x^2} \) can be simplified by algebraic division.

Question 9. ∫ x sin¯¹ x / \( \sqrt{1-x^2} \) dx
Answer:
For \( \int \frac{x \sin^{-1} x}{\sqrt{1-x^2}} \, dx \), we use substitution.
Let \( t = \sin^{-1} x \). Then \( \frac{dt}{dx} = \frac{1}{\sqrt{1-x^2}} \implies dt = \frac{1}{\sqrt{1-x^2}} \, dx \).
Also, from \( t = \sin^{-1} x \), we have \( x = \sin t \).
The integral becomes \( \int (\sin t) (t) \, dt = \int t \sin t \, dt \).
Now, apply integration by parts to \( \int t \sin t \, dt \). Let \( u = t \) (first function) and \( dv = \sin t \, dt \) (second function). Then \( du = 1 \, dt \) and \( v = -\cos t \).
\( \int t \sin t \, dt = t (-\cos t) - \int (1) (-\cos t) \, dt \)
\( = -t \cos t + \int \cos t \, dt \)
\( = -t \cos t + \sin t + C \).
Substitute back \( t = \sin^{-1} x \). Since \( x = \sin t \), then \( \cos t = \sqrt{1-\sin^2 t} = \sqrt{1-x^2} \).
\( = -\sin^{-1} x \cdot \sqrt{1-x^2} + x + C \).
In simple words: We first use a substitution for \( \sin^{-1} x \) to make the integral simpler. This turns the problem into a basic integration by parts problem involving \( t \sin t \). After solving that, we switch back to the original variable \( x \). This technique simplifies the problem into a known form.

🎯 Exam Tip: When you see \( \sin^{-1} x \) along with \( \frac{1}{\sqrt{1-x^2}} \), a substitution \( t = \sin^{-1} x \) is usually the best first step. Similarly, for \( \tan^{-1} x \) with \( \frac{1}{1+x^2} \), use \( t = \tan^{-1} x \).

Question 10. ∫ x tan¯¹ x / (1+x²)\(^{3/2}\) dx
Answer:
For \( \int \frac{x \tan^{-1} x}{(1+x^2)^{3/2}} \, dx \), we use a trigonometric substitution.
Let \( x = \tan \theta \). Then \( dx = \sec^2 \theta \, d\theta \).
Also, \( \tan^{-1} x = \theta \).
\( (1+x^2)^{3/2} = (1+\tan^2 \theta)^{3/2} = (\sec^2 \theta)^{3/2} = \sec^3 \theta \).
Substitute these into the integral:
\( \int \frac{(\tan \theta)(\theta)}{\sec^3 \theta} (\sec^2 \theta \, d\theta) = \int \frac{\theta \tan \theta}{\sec \theta} \, d\theta \)
\( = \int \theta \frac{\sin \theta}{\cos \theta} \cos \theta \, d\theta = \int \theta \sin \theta \, d\theta \).
Now, apply integration by parts to \( \int \theta \sin \theta \, d\theta \). Let \( u = \theta \) (first function) and \( dv = \sin \theta \, d\theta \) (second function). Then \( du = 1 \, d\theta \) and \( v = -\cos \theta \).
\( \int \theta \sin \theta \, d\theta = \theta (-\cos \theta) - \int (1) (-\cos \theta) \, d\theta \)
\( = -\theta \cos \theta + \int \cos \theta \, d\theta \)
\( = -\theta \cos \theta + \sin \theta + C \).
Substitute back \( \theta = \tan^{-1} x \). If \( x = \tan \theta \), then we can draw a right triangle with opposite side \( x \) and adjacent side \( 1 \), so the hypotenuse is \( \sqrt{1+x^2} \).
Thus, \( \sin \theta = \frac{x}{\sqrt{1+x^2}} \) and \( \cos \theta = \frac{1}{\sqrt{1+x^2}} \).
Substitute these back:
\( = -(\tan^{-1} x) \left( \frac{1}{\sqrt{1+x^2}} \right) + \frac{x}{\sqrt{1+x^2}} + C \)
\( = \frac{x - \tan^{-1} x}{\sqrt{1+x^2}} + C \).
In simple words: This problem becomes much easier by changing the variable from \( x \) to \( \tan\theta \). This simplifies the complex fraction and the inverse tangent part. After that, we use integration by parts to solve the new integral, then change back to \( x \) using a right-angle triangle.

🎯 Exam Tip: For integrals involving \( (a^2+x^2)^{3/2} \) or \( \tan^{-1}x \), the substitution \( x=a\tan\theta \) is very common and effective. Remember to convert \( \sin\theta \) and \( \cos\theta \) back to \( x \) using a right-angle triangle from the original substitution.

Question 12. ∫ (2x + sin 2x) / (1+cos 2x) dx
Answer:
For \( \int \frac{2x + \sin 2x}{1+\cos 2x} \, dx \), we use trigonometric identities to simplify the expression.
We know \( \sin 2x = 2 \sin x \cos x \) and \( 1+\cos 2x = 2\cos^2 x \).
Substitute these into the integral:
\( \int \frac{2x + 2 \sin x \cos x}{2\cos^2 x} \, dx \)
Split the fraction into two parts:
\( = \int \left( \frac{2x}{2\cos^2 x} + \frac{2 \sin x \cos x}{2\cos^2 x} \right) \, dx \)
\( = \int \left( \frac{x}{\cos^2 x} + \frac{\sin x}{\cos x} \right) \, dx \)
\( = \int (x \sec^2 x + \tan x) \, dx \)
\( = \int x \sec^2 x \, dx + \int \tan x \, dx \)
For \( \int x \sec^2 x \, dx \), we use integration by parts. Let \( u = x \) (first function) and \( dv = \sec^2 x \, dx \) (second function). Then \( du = 1 \, dx \) and \( v = \tan x \).
\( \int x \sec^2 x \, dx = x \tan x - \int (1) \tan x \, dx \)
\( = x \tan x - \int \tan x \, dx \)
Substitute this back into the main expression:
\( = (x \tan x - \int \tan x \, dx) + \int \tan x \, dx + C \)
The two \( \int \tan x \, dx \) terms cancel out.
\( = x \tan x + C \).
In simple words: We first use known rules for sine and cosine of double angles to simplify the fraction. This turns the integral into two simpler parts. One part is solved using integration by parts, and the other part cancels out, leaving a clean final answer.

🎯 Exam Tip: Always look for opportunities to simplify trigonometric expressions using identities, especially for double angles, before attempting integration. This often reduces complex problems to much simpler forms.

Question 13. ∫ eˣ (1-sin x)/(1-cos x) dx
Answer:
For \( \int e^x \frac{1-\sin x}{1-\cos x} \, dx \), we use trigonometric half-angle identities to simplify the fraction.
We know \( \sin x = 2 \sin(x/2) \cos(x/2) \) and \( 1-\cos x = 2\sin^2(x/2) \).
Substitute these into the fraction:
\( \frac{1-\sin x}{1-\cos x} = \frac{1 - 2 \sin(x/2) \cos(x/2)}{2\sin^2(x/2)} \)
Split the fraction into two parts:
\( = \frac{1}{2\sin^2(x/2)} - \frac{2 \sin(x/2) \cos(x/2)}{2\sin^2(x/2)} \)
\( = \frac{1}{2} \operatorname{cosec}^2(x/2) - \frac{\cos(x/2)}{\sin(x/2)} \)
\( = \frac{1}{2} \operatorname{cosec}^2(x/2) - \cot(x/2) \).
So the integral becomes \( \int e^x \left( \frac{1}{2} \operatorname{cosec}^2(x/2) - \cot(x/2) \right) \, dx \).
This is of the form \( \int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C \).
Let \( f(x) = -\cot(x/2) \). Then, using the chain rule, \( f'(x) = -(-\operatorname{cosec}^2(x/2)) \cdot \frac{1}{2} = \frac{1}{2} \operatorname{cosec}^2(x/2) \).
So, the integral matches the form \( \int e^x (f(x) + f'(x)) \, dx \).
Therefore, the solution is \( e^x (-\cot(x/2)) + C = -e^x \cot(x/2) + C \).
In simple words: We use half-angle trigonometric formulas to break down the complicated fraction. This transforms the integral into a special form where the answer is simply \( e^x \) multiplied by one of the terms inside the parenthesis. This is a very handy trick for specific types of integrals involving \( e^x \).

🎯 Exam Tip: Always be on the lookout for the \( \int e^x (f(x) + f'(x)) \, dx \) pattern when \( e^x \) is multiplied by a sum of functions. Simplifying the expression often reveals this form, making the integration straightforward.

Question 14. ∫eˣ (log x + 1/x²) dx
Answer:
For \( \int e^x \left( \log x + \frac{1}{x^2} \right) \, dx \), we can split this into two integrals and use integration by parts for one of them.
\( I = \int e^x \log x \, dx + \int e^x \frac{1}{x^2} \, dx \)
Let's apply integration by parts to the first integral, \( \int e^x \log x \, dx \). Let \( u = \log x \) (first function) and \( dv = e^x \, dx \) (second function). Then \( du = \frac{1}{x} \, dx \) and \( v = e^x \).
\( \int e^x \log x \, dx = (\log x) e^x - \int \frac{1}{x} e^x \, dx \)
Now, let's apply integration by parts again to the new integral \( \int \frac{1}{x} e^x \, dx \). Let \( u = \frac{1}{x} \) (first function) and \( dv = e^x \, dx \) (second function). Then \( du = -\frac{1}{x^2} \, dx \) and \( v = e^x \).
\( \int \frac{1}{x} e^x \, dx = \left( \frac{1}{x} \right) e^x - \int \left( -\frac{1}{x^2} \right) e^x \, dx \)
\( = \frac{e^x}{x} + \int \frac{e^x}{x^2} \, dx \)
Substitute this back into the expression for \( \int e^x \log x \, dx \):
\( \int e^x \log x \, dx = e^x \log x - \left( \frac{e^x}{x} + \int \frac{e^x}{x^2} \, dx \right) \)
\( = e^x \log x - \frac{e^x}{x} - \int \frac{e^x}{x^2} \, dx \)
Now, substitute this back into the original full integral \( I \):
\( I = \left( e^x \log x - \frac{e^x}{x} - \int \frac{e^x}{x^2} \, dx \right) + \int e^x \frac{1}{x^2} \, dx + C \)
The two integral terms \( -\int \frac{e^x}{x^2} \, dx \) and \( +\int e^x \frac{1}{x^2} \, dx \) cancel each other out.
\( I = e^x \log x - \frac{e^x}{x} + C \)
\( = e^x \left( \log x - \frac{1}{x} \right) + C \).
In simple words: This integral is solved by a clever use of integration by parts. We apply the rule to the \( e^x \log x \) part. This creates a new integral that neatly cancels out the other term in the original problem, leaving a simple final answer. It's a useful trick to remember when terms seem to cancel.

🎯 Exam Tip: This integral is a variation of the \( \int e^x (f(x) + f'(x)) \, dx \) form. If you identify \( f(x) = \log x - \frac{1}{x} \), then \( f'(x) = \frac{1}{x} + \frac{1}{x^2} \), which is not exactly the form. However, if you treat the first integral \( \int e^x \log x \, dx \) using parts, the resulting \( \int \frac{e^x}{x} \, dx \) is then treated with parts to yield a term that cancels the \( \frac{e^x}{x^2} \) part. Keep an eye out for such cancellations after repeated integration by parts.

Question 15. \( \int e^{x} [\log (\sec x + \tan x) + \sec x] dx \)
Answer: This integral is in the form \( \int e^{x} [f(x) + f'(x)] dx \), which equals \( e^{x} f(x) + C \).
Here, let \( f(x) = \log (\sec x + \tan x) \).
Now, we find the derivative of \( f(x) \):
\( f'(x) = \frac{d}{dx} [\log (\sec x + \tan x)] \)
\( f'(x) = \frac{1}{(\sec x + \tan x)} \cdot \frac{d}{dx} (\sec x + \tan x) \)
\( f'(x) = \frac{1}{(\sec x + \tan x)} \cdot (\sec x \tan x + \sec^{2} x) \)
\( f'(x) = \frac{\sec x (\tan x + \sec x)}{(\sec x + \tan x)} \)
\( f'(x) = \sec x \)
Since the integral matches the form \( \int e^{x} [f(x) + f'(x)] dx \), the solution is:
\( \int e^{x} [\log (\sec x + \tan x) + \sec x] dx = e^{x} \log (\sec x + \tan x) + C \)
This type of integral simplifies nicely when one part is the derivative of the other.
In simple words: This problem uses a special rule for integrals where you have \( e^{x} \) multiplied by a function plus its derivative. If you can spot this pattern, the answer is simply \( e^{x} \) multiplied by the original function.

🎯 Exam Tip: Always look for the \( \int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C \) pattern in integrals involving \( e^{x} \). Recognizing \( f(x) \) and \( f'(x) \) correctly is key.

Question 16. \( \int e^{x} (\sin x + \cos x) \sec^{2} x dx \)
Answer: We need to rearrange the expression inside the integral to fit the form \( \int e^{x} [f(x) + f'(x)] dx \).
The term is \( (\sin x + \cos x) \sec^{2} x \).
Let's expand it:
\( \sin x \sec^{2} x + \cos x \sec^{2} x \)
\( = \sin x \cdot \frac{1}{\cos^{2} x} + \cos x \cdot \frac{1}{\cos^{2} x} \)
\( = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} + \frac{1}{\cos x} \)
\( = \tan x \sec x + \sec x \)
So the integral becomes \( \int e^{x} (\sec x + \tan x \sec x) dx \).
Here, let \( f(x) = \sec x \).
The derivative of \( f(x) \) is \( f'(x) = \frac{d}{dx} (\sec x) = \sec x \tan x \).
Thus, the integral is in the form \( \int e^{x} [f(x) + f'(x)] dx \).
The solution is \( e^{x} f(x) + C \).
\( = e^{x} \sec x + C \)
It's helpful to simplify trigonometric expressions to find hidden patterns.
In simple words: First, break down the complex part of the integral into two simpler terms. Then, identify a function and its derivative among these terms. Once you see the pattern \( e^{x} \) times (function plus its derivative), the answer is simply \( e^{x} \) times that function.

🎯 Exam Tip: When dealing with trigonometric functions inside an \( e^{x} \) integral, simplify the expressions using identities to reveal the \( f(x) + f'(x) \) pattern.

Question 17. \( \int e^{x} (\frac{1}{x^{2}} - \frac{2}{x^{3}}) dx \)
Answer: This integral is in the form \( \int e^{x} [f(x) + f'(x)] dx \).
Let \( f(x) = \frac{1}{x^{2}} \).
We can write \( f(x) \) as \( x^{-2} \).
Now, let's find its derivative, \( f'(x) \):
\( f'(x) = \frac{d}{dx} (x^{-2}) \)
Using the power rule, \( f'(x) = -2x^{-2-1} = -2x^{-3} \)
So, \( f'(x) = -\frac{2}{x^{3}} \).
The given integral is \( \int e^{x} (\frac{1}{x^{2}} - \frac{2}{x^{3}}) dx \), which exactly matches \( \int e^{x} [f(x) + f'(x)] dx \).
Therefore, the solution is \( e^{x} f(x) + C \).
\( = e^{x} \cdot \frac{1}{x^{2}} + C \)
This type of problem demonstrates a fundamental property of integrals involving exponential functions.
In simple words: This is a special kind of integral. If you have \( e^{x} \) multiplied by a function and its derivative added together, the answer is just \( e^{x} \) multiplied by the original function. Here, \( 1/x^{2} \) is the function and \( -2/x^{3} \) is its derivative.

🎯 Exam Tip: Always check if an integral involving \( e^{x} \) can be written as \( \int e^{x} [f(x) + f'(x)] dx \). This shortcut saves a lot of time in complex problems.

Question 18. \( \int e^{x} (\frac{1-x}{1+x^{2}})^{2} dx \)
Answer: We need to simplify the expression inside the integral to fit the form \( \int e^{x} [f(x) + f'(x)] dx \).
Let \( I = \int e^{x} \left( \frac{1-x}{1+x^{2}} \right)^{2} dx \).
First, expand the numerator:
\( \left( \frac{1-x}{1+x^{2}} \right)^{2} = \frac{(1-x)^{2}}{(1+x^{2})^{2}} = \frac{1 - 2x + x^{2}}{(1+x^{2})^{2}} \)
Now, rewrite the numerator to create separate terms:
\( = \frac{(1+x^{2}) - 2x}{(1+x^{2})^{2}} \)
Separate the fraction:
\( = \frac{(1+x^{2})}{(1+x^{2})^{2}} - \frac{2x}{(1+x^{2})^{2}} \)
\( = \frac{1}{1+x^{2}} - \frac{2x}{(1+x^{2})^{2}} \)
Now, the integral becomes \( I = \int e^{x} \left( \frac{1}{1+x^{2}} - \frac{2x}{(1+x^{2})^{2}} \right) dx \).
Let \( f(x) = \frac{1}{1+x^{2}} \).
To find \( f'(x) \), we use the chain rule (or quotient rule). Let \( u = 1 \) and \( v = 1+x^{2} \).
\( f'(x) = \frac{v u' - u v'}{v^{2}} = \frac{(1+x^{2})(0) - (1)(2x)}{(1+x^{2})^{2}} = -\frac{2x}{(1+x^{2})^{2}} \).
This matches the second term in the integral, \( -\frac{2x}{(1+x^{2})^{2}} \).
So, the integral is in the form \( \int e^{x} [f(x) + f'(x)] dx \).
The solution is \( e^{x} f(x) + C \).
\( I = e^{x} \cdot \frac{1}{1+x^{2}} + C \)
Recognizing this pattern is a key skill for solving such integrals efficiently.
In simple words: The trick here is to rewrite the complex fraction so it looks like a function plus its derivative. We expanded the top part and then split the fraction into two. One part became \( 1/(1+x^{2}) \), and the other part was exactly its derivative. So, the final answer is \( e^{x} \) multiplied by \( 1/(1+x^{2}) \).

🎯 Exam Tip: When faced with a squared term in the denominator like \( (1+x^{2})^{2} \), try manipulating the numerator to create terms that can be separated and simplified, often revealing the \( f(x) + f'(x) \) pattern.

Question 20. \( \int \frac{x^{2}}{(x \cos x - \sin x)^{2}} dx \)
Answer: Let's follow the provided solution steps for this integral. The solution seems to address the integral \( I = \int \frac{x \sin x}{(x \cos x - \sin x)^{2}} dx \). We will proceed with solving this integral based on the steps given.
Let \( u = x \cos x - \sin x \).
Then, find the differential \( du \):
\( du = \frac{d}{dx} (x \cos x - \sin x) dx \)
Using the product rule for \( x \cos x \): \( \frac{d}{dx} (x \cos x) = 1 \cdot \cos x + x (-\sin x) = \cos x - x \sin x \).
The derivative of \( \sin x \) is \( \cos x \).
So, \( du = (\cos x - x \sin x - \cos x) dx \)
\( du = (-x \sin x) dx \)
This means \( x \sin x dx = -du \).
Substitute \( u \) and \( du \) into the integral \( I \):
\( I = \int \frac{-du}{u^{2}} \)
\( I = -\int u^{-2} du \)
Now, integrate \( u^{-2} \):
\( I = - \left( \frac{u^{-2+1}}{-2+1} \right) + C \)
\( I = - \left( \frac{u^{-1}}{-1} \right) + C \)
\( I = - (-u^{-1}) + C \)
\( I = \frac{1}{u} + C \)
Finally, substitute back \( u = x \cos x - \sin x \):
\( I = \frac{1}{x \cos x - \sin x} + C \)
This method of substitution simplifies the integral effectively. Always ensure that the substitution covers all parts of the original integral.
In simple words: For this type of integral, we use a trick called substitution. We let the bottom part, \( x \cos x - \sin x \), be a new variable, \( u \). Then we find the derivative of \( u \) to replace the \( x \sin x \, dx \) part. This changes the problem into a much simpler integral of \( 1/u^{2} \), which is easy to solve. Finally, we put the original expression back in place of \( u \).

🎯 Exam Tip: When you see a function and its derivative (or a multiple of it) in an integral, consider substitution. For fractions with powers, try substituting the denominator or a part of it. This often simplifies the integral dramatically.

Question 21. \( I = \int \cos^{-1} \left(\frac{1}{x}\right) dx \)
Answer: We first simplify the inverse trigonometric function.
We know that \( \cos^{-1} \left(\frac{1}{x}\right) = \sec^{-1} x \).
So the integral becomes \( I = \int \sec^{-1} x dx \).
To solve this, we use integration by parts, treating it as \( \int \sec^{-1} x \cdot 1 dx \).
Let \( u = \sec^{-1} x \) and \( dv = 1 dx \).
Then, \( du = \frac{1}{|x|\sqrt{x^{2}-1}} dx \) (or \( \frac{1}{x\sqrt{x^{2}-1}} dx \) for \( x>1 \)) and \( v = \int 1 dx = x \).
The integration by parts formula is \( \int u dv = uv - \int v du \).
\( I = x \sec^{-1} x - \int x \cdot \frac{1}{x\sqrt{x^{2}-1}} dx \)
\( I = x \sec^{-1} x - \int \frac{1}{\sqrt{x^{2}-1}} dx \)
The integral \( \int \frac{1}{\sqrt{x^{2}-a^{2}}} dx = \log |x + \sqrt{x^{2}-a^{2}}| + C \). Here \( a=1 \).
So, \( I = x \sec^{-1} x - \log |x + \sqrt{x^{2}-1}| + C \)
This integral combines an inverse trigonometric identity with the method of integration by parts.
In simple words: First, we change \( \cos^{-1}(1/x) \) into \( \sec^{-1} x \) because they are the same. Then, we use a method called "integration by parts" by treating the integral as \( \sec^{-1} x \) multiplied by 1. We find the derivative of \( \sec^{-1} x \) and integrate 1, then put them into the formula. Finally, we solve a standard integral of \( 1/\sqrt{x^{2}-1} \) to get the full answer.

🎯 Exam Tip: Always look for opportunities to simplify inverse trigonometric functions using identities like \( \cos^{-1}(1/x) = \sec^{-1} x \) before applying integration by parts, as it can make the process much smoother.

Question 22. \( \int (\sin^{-1} x)^{2} dx \)
Answer: Let \( I = \int (\sin^{-1} x)^{2} dx \).
We use substitution to simplify the integral. Let \( \sin^{-1} x = \theta \).
This implies \( x = \sin \theta \).
Now, find \( dx \) by differentiating \( x \) with respect to \( \theta \):
\( dx = \cos \theta d\theta \).
Substitute these into the integral:
\( I = \int \theta^{2} \cos \theta d\theta \).
This is an integral that requires integration by parts, applied twice.
First application of integration by parts: \( \int u dv = uv - \int v du \).
Let \( u = \theta^{2} \) and \( dv = \cos \theta d\theta \).
Then, \( du = 2\theta d\theta \) and \( v = \int \cos \theta d\theta = \sin \theta \).
So, \( I = \theta^{2} \sin \theta - \int \sin \theta (2\theta d\theta) \)
\( I = \theta^{2} \sin \theta - 2 \int \theta \sin \theta d\theta \).
Now, we need to solve \( \int \theta \sin \theta d\theta \) using integration by parts again.
Let \( u' = \theta \) and \( dv' = \sin \theta d\theta \).
Then, \( du' = d\theta \) and \( v' = \int \sin \theta d\theta = -\cos \theta \).
So, \( \int \theta \sin \theta d\theta = \theta (-\cos \theta) - \int (-\cos \theta) d\theta \)
\( = -\theta \cos \theta + \int \cos \theta d\theta \)
\( = -\theta \cos \theta + \sin \theta \).
Substitute this result back into the expression for \( I \):
\( I = \theta^{2} \sin \theta - 2 (-\theta \cos \theta + \sin \theta) + C \)
\( I = \theta^{2} \sin \theta + 2\theta \cos \theta - 2 \sin \theta + C \).
Finally, substitute back \( \theta = \sin^{-1} x \), \( \sin \theta = x \), and \( \cos \theta = \sqrt{1 - \sin^{2} \theta} = \sqrt{1 - x^{2}} \).
\( I = (\sin^{-1} x)^{2} x + 2(\sin^{-1} x) \sqrt{1 - x^{2}} - 2x + C \)
This problem shows how a combination of substitution and repeated integration by parts can solve complex integrals. Breaking down the problem into smaller, manageable steps is very effective.
In simple words: First, we change the \( \sin^{-1} x \) to \( \theta \) and replace \( x \) with \( \sin \theta \). This changes the integral into \( \int \theta^{2} \cos \theta d\theta \). We then solve this new integral using "integration by parts" not once, but twice, because there's still a product of two functions. After solving, we put the original \( \sin^{-1} x \) back in place of \( \theta \) to get the final answer.

🎯 Exam Tip: Integrals of inverse trigonometric functions often become easier with a substitution like \( u = \sin^{-1} x \) or \( u = \tan^{-1} x \). This transforms the integral into a product of algebraic and trigonometric functions, which can then be solved using integration by parts (often multiple times).