GSEB Class 12 Maths Solutions Chapter 7 Integrals Exercise 7.6

Integrate the Following Functions:

Question 1. x sin x
Answer: Let \( I = \int x \sin x \, dx \).
We know that the integration by parts formula is \( \int uv \, dx = u \int v \, dx - \int (\frac{du}{dx} \int v \, dx) \, dx \).
Put \( u = x \) and \( v = \sin x \).
So, \( I = x \int \sin x \, dx - \int (\frac{d}{dx} x \int \sin x \, dx) \, dx \)
\( = -x \cos x - \int 1 \cdot (-\cos x) \, dx \)
\( = -x \cos x + \int \cos x \, dx \)
\( = -x \cos x + \sin x + C \).
In simple words: To solve this, we use a special method called integration by parts. We picked 'x' as one part and 'sin x' as the other. After doing the steps, the answer came out as negative x times cos x, plus sin x, plus a constant 'C'.

Exam Tip: Remember the LIATE rule (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) to choose 'u' effectively for integration by parts.

Question 2. x sin 3x
Answer: Let \( I = \int x \sin 3x \, dx \). Taking \( u = x \) and \( v = \sin 3x \).
Now, \( \int uv \, dx = u \int v \, dx - \int (\frac{du}{dx} \int v \, dx) \, dx \)
So, \( I = x \int \sin 3x \, dx - \int (\frac{d}{dx} x \int \sin 3x \, dx) \, dx \)
\( = x (-\frac{\cos 3x}{3}) - \int (1 \cdot (-\frac{\cos 3x}{3})) \, dx \)
\( = -\frac{x \cos 3x}{3} + \frac{1}{3} \int \cos 3x \, dx \)
\( = -\frac{x \cos 3x}{3} + \frac{1}{3} (\frac{\sin 3x}{3}) + C \)
\( = -\frac{x \cos 3x}{3} + \frac{\sin 3x}{9} + C \).
In simple words: We used integration by parts again. We chose 'x' as our first term and 'sin 3x' as our second. After integrating and simplifying, the final result involved x times cos 3x, and sin 3x, plus the constant.

Exam Tip: When integrating functions like `sin(ax+b)` or `cos(ax+b)`, remember to divide by the coefficient of `x` (i.e., `a`).

Question 3. x²ex
Answer: Let \( I = \int x^2 e^x \, dx \). Put \( u = x^2 \) and \( v = e^x dx \).
From \( \int uv \, dx = u \int v \, dx - \int (\frac{du}{dx} \int v \, dx) \, dx \)
\( I = x^2 \int e^x \, dx - \int (2x \cdot e^x) \, dx \)
\( = x^2 e^x - 2 \int x e^x \, dx \).
Taking x as I function, integrating again, we get
\( I = x^2 e^x - 2 [x \int e^x \, dx - \int (\frac{d}{dx} x \int e^x \, dx) \, dx] \)
\( = x^2 e^x - 2 [x e^x - \int (1 \cdot e^x) \, dx] \)
\( = x^2 e^x - 2 [x e^x - e^x] + C \)
\( = x^2 e^x - 2x e^x + 2e^x + C \)
\( = e^x(x^2 - 2x + 2) + C \).
In simple words: This problem required using the integration by parts rule two times. First, we chose \(x^2\) as 'u'. After one round, we still had \(x e^x\) to integrate, so we applied the rule again with 'x' as 'u'. Finally, we gathered all the terms to get our answer.

Exam Tip: When dealing with \(x^n e^x\) type integrals, you might need to apply integration by parts 'n' times. Be meticulous with each step.

Question 4. x log x
Answer: Let \( I = \int x \log x \, dx \). Put \( u = \log x \) and \( v = x \).
We know that \( \int uv \, dx = u \int v \, dx - \int (\frac{du}{dx} \int v \, dx) \, dx \).
So, \( \int (\log x) x \, dx = \log(x) \int x \, dx - \int (\frac{d}{dx} (\log x) \int x \, dx) \, dx \)
\( = (\log x) \frac{x^2}{2} - \int (\frac{1}{x} \cdot \frac{x^2}{2}) \, dx \)
\( = \frac{x^2}{2} \log x - \frac{1}{2} \int x \, dx \)
\( = \frac{x^2}{2} \log x - \frac{x^2}{4} + C \).
In simple words: For this integral, we used the integration by parts method. We chose \( \log x \) as the 'u' term because it simplifies when differentiated. After working through the formula, we got a term with \( \log x \) and another term with \( x^2 \), plus the constant.

Exam Tip: For products of algebraic and logarithmic functions, always pick the logarithmic function as 'u' when using integration by parts, as its derivative is simpler.

Question 5. x log 2x
Answer: Let \( I = \int x \log 2x \, dx = \int (\log 2x) \cdot x \, dx \).
\( = (\log 2x) \cdot \frac{x^2}{2} - \int (\frac{1}{2x} \cdot 2 \cdot \frac{x^2}{2}) \, dx \)
\( = \frac{x^2}{2} \log 2x - \int \frac{x}{2} \, dx \)
\( = \frac{x^2}{2} \log 2x - \frac{1}{2} \cdot \frac{x^2}{2} + C \)
\( = \frac{x^2}{2} \log 2x - \frac{x^2}{4} + C \).
In simple words: We used integration by parts for this problem. We picked \( \log 2x \) as 'u' and 'x' as 'dv'. After doing the steps and simplifying, the final answer was \( \frac{x^2}{2} \log 2x - \frac{x^2}{4} + C \).

Exam Tip: The derivative of \( \log(ax) \) is \( 1/x \), just like \( \log x \). This can simplify calculations during integration by parts.

Question 6. x²log x
Answer: Let \( I = \int x^2 \log x \, dx \).
Using integration by parts, with \( u = \log|x| \) and \( dv = x^2 \, dx \).
Then \( du = \frac{1}{x} \, dx \) and \( v = \frac{x^3}{3} \).
So, \( I = (\log|x|) \frac{x^3}{3} - \int (\frac{x^3}{3} \cdot \frac{1}{x}) \, dx \)
\( = \frac{x^3}{3} \log|x| - \frac{1}{3} \int x^2 \, dx \)
\( = \frac{x^3}{3} \log|x| - \frac{1}{3} \cdot \frac{x^3}{3} + C \)
\( = \frac{x^3}{3} \log|x| - \frac{x^3}{9} + C \).
In simple words: This integral needed the integration by parts method. We chose \( \log x \) as 'u' and \( x^2 \) as 'dv' because it simplifies the logarithm. After applying the rule and doing the required integrations, we got the final expression involving \( x^3 \log x \) and \( x^3 \).

Exam Tip: Remember to use \( \log|x| \) in indefinite integrals involving \( \log x \) to ensure the domain is correct, especially if \( x \) can be negative.

Question 7. x sin⁻¹x
Answer: Let \( I = \int x \sin^{-1} x \, dx = \int \sin^{-1} x \cdot x \, dx \).
Using integration by parts with \( u = \sin^{-1} x \) and \( dv = x \, dx \).
So, \( du = \frac{1}{\sqrt{1-x^2}} \, dx \) and \( v = \frac{x^2}{2} \).
\( I = \sin^{-1} x \cdot \frac{x^2}{2} - \int (\frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2}) \, dx \)
\( = \frac{x^2}{2} \sin^{-1} x - \frac{1}{2} \int \frac{x^2}{\sqrt{1-x^2}} \, dx \).
Let \( I_1 = \int \frac{x^2}{\sqrt{1-x^2}} \, dx \).
Put \( x = \sin \theta \) so that \( dx = \cos \theta \, d\theta \).
\( I_1 = \int \frac{\sin^2 \theta}{\sqrt{1-\sin^2 \theta}} \cos \theta \, d\theta = \int \frac{\sin^2 \theta}{\cos \theta} \cos \theta \, d\theta = \int \sin^2 \theta \, d\theta \)
\( = \int \frac{1-\cos 2\theta}{2} \, d\theta = \frac{1}{2} \int (1 - \cos 2\theta) \, d\theta \)
\( = \frac{1}{2} [\theta - \frac{\sin 2\theta}{2}] + C_1 \)
\( = \frac{1}{2} [\theta - \frac{2 \sin \theta \cos \theta}{2}] + C_1 = \frac{1}{2} [\theta - \sin \theta \cos \theta] + C_1 \).
Substitute back: \( \theta = \sin^{-1} x \). Since \( x = \sin \theta \), then \( \cos \theta = \sqrt{1-\sin^2 \theta} = \sqrt{1-x^2} \).
\( I_1 = \frac{1}{2} [\sin^{-1} x - x\sqrt{1-x^2}] + C_1 \).
Now, substitute \( I_1 \) back into the expression for \( I \):
\( I = \frac{x^2}{2} \sin^{-1} x - \frac{1}{2} \cdot \frac{1}{2} [\sin^{-1} x - x\sqrt{1-x^2}] + C \)
\( = \frac{x^2}{2} \sin^{-1} x - \frac{1}{4} \sin^{-1} x + \frac{x\sqrt{1-x^2}}{4} + C \)
\( = \sin^{-1} x (\frac{x^2}{2} - \frac{1}{4}) + \frac{x\sqrt{1-x^2}}{4} + C \)
\( = \frac{1}{4} \sin^{-1} x (2x^2 - 1) + \frac{x\sqrt{1-x^2}}{4} + C \).
In simple words: This problem was solved using integration by parts, where \( \sin^{-1}x \) was 'u'. The resulting integral needed another substitution, \( x = \sin \theta \), to simplify. After performing that substitution and integrating, we changed back to 'x' and combined all the parts for the final answer.

Exam Tip: Integrals involving inverse trigonometric functions often require integration by parts, and then a trigonometric substitution for the remaining integral part.

Question 8. x tan⁻¹x
Answer: Let \( I = \int x \tan^{-1} x \, dx \). Put \( u = \tan^{-1} x \) and \( v = x \).
We know that \( \int uv \, dx = u \int v \, dx - \int (\frac{du}{dx} \int v \, dx) \, dx \).
So, \( \int (\tan^{-1} x) \cdot x \, dx = \tan^{-1} x \int x \, dx - \int (\frac{d}{dx} (\tan^{-1} x) \int x \, dx) \, dx \)
\( = \tan^{-1} x \cdot \frac{x^2}{2} - \int (\frac{1}{1+x^2} \cdot \frac{x^2}{2}) \, dx \)
\( = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int \frac{x^2}{1+x^2} \, dx \)
\( = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int \frac{x^2+1-1}{x^2+1} \, dx \)
\( = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int (1 - \frac{1}{x^2+1}) \, dx \)
\( = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} [x - \tan^{-1} x] + C \)
\( = \frac{x^2}{2} \tan^{-1} x - \frac{x}{2} + \frac{1}{2} \tan^{-1} x + C \)
\( = \frac{x^2+1}{2} \tan^{-1} x - \frac{x}{2} + C \).
In simple words: To integrate \( x \tan^{-1}x \), we used the integration by parts rule. We chose \( \tan^{-1}x \) as 'u' and 'x' as 'dv'. After applying the rule, we had to simplify a fraction and integrate it, which gave us the final answer.

Exam Tip: When faced with \( \int \frac{x^2}{1+x^2} \, dx \), remember the trick of adding and subtracting 1 in the numerator to simplify the integrand.

Question 9. x cos⁻¹x
Answer: Let \( I = \int x \cos^{-1} x \, dx \).
Using integration by parts, with \( u = \cos^{-1} x \) and \( dv = x \, dx \).
So, \( du = -\frac{1}{\sqrt{1-x^2}} \, dx \) and \( v = \frac{x^2}{2} \).
\( I = \cos^{-1} x \cdot \frac{x^2}{2} - \int (-\frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2}) \, dx \)
\( = \frac{x^2}{2} \cos^{-1} x + \frac{1}{2} \int \frac{x^2}{\sqrt{1-x^2}} \, dx \).
Let \( I_1 = \int \frac{x^2}{\sqrt{1-x^2}} \, dx \).
Put \( x = \cos \theta \) so that \( dx = -\sin \theta \, d\theta \).
\( I_1 = \int \frac{\cos^2 \theta}{\sqrt{1-\cos^2 \theta}} (-\sin \theta) \, d\theta = \int \frac{\cos^2 \theta}{\sin \theta} (-\sin \theta) \, d\theta = -\int \cos^2 \theta \, d\theta \)
\( = -\int \frac{1+\cos 2\theta}{2} \, d\theta = -\frac{1}{2} \int (1+\cos 2\theta) \, d\theta \)
\( = -\frac{1}{2} [\theta + \frac{\sin 2\theta}{2}] + C_1 = -\frac{1}{2} [\theta + \sin \theta \cos \theta] + C_1 \).
Substitute back: \( \theta = \cos^{-1} x \). Since \( x = \cos \theta \), then \( \sin \theta = \sqrt{1-\cos^2 \theta} = \sqrt{1-x^2} \).
\( I_1 = -\frac{1}{2} [\cos^{-1} x + x\sqrt{1-x^2}] + C_1 \).
Now, substitute \( I_1 \) back into the expression for \( I \):
\( I = \frac{x^2}{2} \cos^{-1} x + \frac{1}{2} \cdot (-\frac{1}{2}) [\cos^{-1} x + x\sqrt{1-x^2}] + C \)
\( = \frac{x^2}{2} \cos^{-1} x - \frac{1}{4} \cos^{-1} x - \frac{x\sqrt{1-x^2}}{4} + C \)
\( = \cos^{-1} x (\frac{x^2}{2} - \frac{1}{4}) - \frac{x\sqrt{1-x^2}}{4} + C \)
\( = \frac{\cos^{-1} x (2x^2-1)}{4} - \frac{x\sqrt{1-x^2}}{4} + C \).
In simple words: This problem used integration by parts, with \( \cos^{-1}x \) as the 'u' term. The leftover integral needed a trigonometric substitution, \( x = \cos \theta \). After that, we integrated and put everything back in terms of 'x' to get the final answer.

Exam Tip: Be careful with the signs when differentiating inverse cosine. Also, trigonometric substitutions often turn complex algebraic integrals into simpler trigonometric ones.

Question 10. (sin⁻¹x)²
Answer: Let \( I = \int (\sin^{-1} x)^2 \, dx \).
Put \( \sin^{-1} x = \theta \Rightarrow x = \sin \theta \). Then \( dx = \cos \theta \, d\theta \).
So, \( I = \int \theta^2 \cos \theta \, d\theta \).
Using integration by parts, with \( u = \theta^2 \) and \( dv = \cos \theta \, d\theta \).
\( I = \theta^2 (\sin \theta) - \int (2\theta \cdot \sin \theta) \, d\theta \)
\( = \theta^2 \sin \theta - 2 \int \theta \sin \theta \, d\theta \).
Now, integrate \( \int \theta \sin \theta \, d\theta \) using parts again with \( u = \theta \) and \( dv = \sin \theta \, d\theta \).
\( \int \theta \sin \theta \, d\theta = \theta (-\cos \theta) - \int (1 \cdot (-\cos \theta)) \, d\theta \)
\( = -\theta \cos \theta + \int \cos \theta \, d\theta \)
\( = -\theta \cos \theta + \sin \theta + C' \).
Substitute this back into the expression for \( I \):
\( I = \theta^2 \sin \theta - 2 [-\theta \cos \theta + \sin \theta] + C \)
\( = \theta^2 \sin \theta + 2\theta \cos \theta - 2\sin \theta + C \).
Now, substitute back \( \theta = \sin^{-1} x \) and \( \sin \theta = x \). Since \( \cos \theta = \sqrt{1-\sin^2 \theta} = \sqrt{1-x^2} \).
\( I = (\sin^{-1} x)^2 \cdot x + 2(\sin^{-1} x) \sqrt{1-x^2} - 2x + C \).
In simple words: We solved this problem by first using a substitution to turn \( (\sin^{-1}x)^2 \) into an integral with \( \theta^2 \cos \theta \). Then, we applied the integration by parts rule twice. Finally, we put all the \( \theta \) terms back into 'x' to get the last answer.

Exam Tip: When dealing with inverse trig functions squared, a common strategy is substitution to simplify the inner function, followed by repeated integration by parts.

Question 11. \( \frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}} \)
Answer: Let \( I = \int \frac{x \cos^{-1} x}{\sqrt{1-x^2}} \, dx \).
Put \( \cos^{-1} x = t \). Then \( -\frac{1}{\sqrt{1-x^2}} \, dx = dt \).
So, \( \frac{1}{\sqrt{1-x^2}} \, dx = -dt \).
Also, from \( \cos^{-1} x = t \), we get \( x = \cos t \).
Substitute these into the integral:
\( I = \int (\cos t) \cdot t \cdot (-dt) = -\int t \cos t \, dt \).
Now, integrate \( \int t \cos t \, dt \) using integration by parts, with \( u = t \) and \( dv = \cos t \, dt \).
So, \( du = dt \) and \( v = \sin t \).
\( \int t \cos t \, dt = t \sin t - \int \sin t \, dt = t \sin t - (-\cos t) + C_1 = t \sin t + \cos t + C_1 \).
Substitute this back into \( I \):
\( I = -(t \sin t + \cos t) + C \).
Now, substitute back \( t = \cos^{-1} x \).
Since \( t = \cos^{-1} x \), then \( \cos t = x \).
And \( \sin t = \sin(\cos^{-1} x) = \sqrt{1-x^2} \).
So, \( I = -(\cos^{-1} x \cdot \sqrt{1-x^2} + x) + C \)
\( = -x - \sqrt{1-x^2} \cos^{-1} x + C \).
In simple words: This integral was solved by first substituting \( \cos^{-1}x \) with 't'. This transformed the problem into a simpler integral \( - \int t \cos t \, dt \). We then used integration by parts to solve this new integral. Finally, we changed 't' back to \( \cos^{-1}x \) to get the final answer.

Exam Tip: Recognize substitution opportunities immediately when you see an inverse trigonometric function with its derivative present in the integrand. Then, use integration by parts for the new form.

Question 12. x sec²x
Answer: Let \( I = \int x \sec^2 x \, dx \). Put \( u = x \) and \( v = \sec^2 x \).
Using integration by parts formula \( \int uv \, dx = u \int v \, dx - \int (\frac{du}{dx} \int v \, dx) \, dx \).
\( I = x \int \sec^2 x \, dx - \int (\frac{d}{dx} x \int \sec^2 x \, dx) \, dx \)
\( = x \tan x - \int (1 \cdot \tan x) \, dx \)
\( = x \tan x - \int \tan x \, dx \)
\( = x \tan x - (-\log|\cos x|) + C \)
\( = x \tan x + \log|\cos x| + C \).
In simple words: We used integration by parts for this problem. We chose 'x' as 'u' and \( \sec^2x \) as 'dv'. After applying the formula, we got a new integral of \( \tan x \), which we knew how to solve. This led us to the final answer of \( x \tan x + \log|\cos x| + C \).

Exam Tip: Remember the standard integral \( \int \sec^2 x \, dx = \tan x \) and \( \int \tan x \, dx = \log|\sec x| \) or \( -\log|\cos x| \).

Question 13. tan⁻¹x
Answer: Let \( I = \int \tan^{-1} x \, dx \). We can write this as \( \int \tan^{-1} x \cdot 1 \, dx \).
Put \( u = \tan^{-1} x \) and \( v = 1 \).
From \( \int uv \, dx = u \int v \, dx - \int (\frac{du}{dx} \int v \, dx) \, dx \).
\( I = \tan^{-1} x \int 1 \, dx - \int (\frac{d}{dx} (\tan^{-1} x) \int 1 \, dx) \, dx \)
\( = \tan^{-1} x \cdot x - \int (\frac{1}{1+x^2} \cdot x) \, dx \)
\( = x \tan^{-1} x - \int \frac{x}{1+x^2} \, dx \).
For the integral \( \int \frac{x}{1+x^2} \, dx \), let \( t = 1+x^2 \). Then \( dt = 2x \, dx \), so \( x \, dx = \frac{1}{2} dt \).
\( \int \frac{x}{1+x^2} \, dx = \int \frac{1}{t} \cdot \frac{1}{2} dt = \frac{1}{2} \int \frac{1}{t} dt = \frac{1}{2} \log|t| + C_1 \).
Substitute back \( t = 1+x^2 \): \( \frac{1}{2} \log(1+x^2) + C_1 \). (Since \( 1+x^2 > 0 \)).
So, \( I = x \tan^{-1} x - \frac{1}{2} \log(1+x^2) + C \).
In simple words: To integrate \( \tan^{-1}x \), we used integration by parts by treating it as \( \tan^{-1}x \cdot 1 \). We chose \( \tan^{-1}x \) as 'u'. After applying the rule, we had a new integral, which we solved using a substitution. This gave us the final answer.

Exam Tip: When integrating inverse trigonometric functions without an obvious 'x' factor, always consider using integration by parts with '1' as the 'dv' term.

Question 14. x(log x)²
Answer: Let \( I = \int x(\log x)^2 \, dx \).
Using integration by parts, with \( u = (\log x)^2 \) and \( dv = x \, dx \).
So, \( du = 2 \log x \cdot \frac{1}{x} \, dx \) and \( v = \frac{x^2}{2} \).
\( I = (\log x)^2 \cdot \frac{x^2}{2} - \int (\frac{x^2}{2} \cdot 2 \log x \cdot \frac{1}{x}) \, dx \)
\( = \frac{x^2}{2} (\log x)^2 - \int x \log x \, dx \).
Now, we need to integrate \( \int x \log x \, dx \) using integration by parts again.
For \( \int x \log x \, dx \), let \( u_1 = \log x \) and \( dv_1 = x \, dx \).
So, \( du_1 = \frac{1}{x} \, dx \) and \( v_1 = \frac{x^2}{2} \).
\( \int x \log x \, dx = (\log x) \frac{x^2}{2} - \int (\frac{x^2}{2} \cdot \frac{1}{x}) \, dx \)
\( = \frac{x^2}{2} \log x - \frac{1}{2} \int x \, dx \)
\( = \frac{x^2}{2} \log x - \frac{1}{2} \cdot \frac{x^2}{2} + C_1 = \frac{x^2}{2} \log x - \frac{x^2}{4} + C_1 \).
Substitute this back into the expression for \( I \):
\( I = \frac{x^2}{2} (\log x)^2 - [\frac{x^2}{2} \log x - \frac{x^2}{4}] + C \)
\( = \frac{x^2}{2} (\log x)^2 - \frac{x^2}{2} \log x + \frac{x^2}{4} + C \).
In simple words: We solved this integral using the integration by parts rule two times. First, we chose \( (\log x)^2 \) as 'u', which led to a new integral of \( x \log x \). Then, we applied integration by parts a second time to solve \( x \log x \). Finally, we combined all the results to get the full answer.

Exam Tip: When \( (\log x)^n \) appears with an algebraic term, repeated integration by parts is common. Be organized with each application to avoid errors.

Question 15. (x² + 1) log x
Answer: Let \( I = \int (x^2+1) \log x \, dx = \int \log x \cdot (x^2+1) \, dx \).
Using integration by parts, with \( u = \log x \) and \( dv = (x^2+1) \, dx \).
So, \( du = \frac{1}{x} \, dx \) and \( v = \frac{x^3}{3} + x \).
\( I = \log x (\frac{x^3}{3} + x) - \int (\frac{x^3}{3} + x) \frac{1}{x} \, dx \)
\( = (\frac{x^3}{3} + x) \log x - \int (\frac{x^2}{3} + 1) \, dx \)
\( = (\frac{x^3}{3} + x) \log x - (\frac{x^3}{9} + x) + C \).
In simple words: For this integral, we used the integration by parts rule. We chose \( \log x \) as 'u' and \( (x^2+1) \) as 'dv' because the derivative of \( \log x \) is simpler. After applying the formula, we integrated the remaining terms to get the final answer.

Exam Tip: When a polynomial and a logarithmic function are multiplied, always choose the logarithmic function as 'u' in integration by parts for easier calculation.

Question 16. ex(sin x + cos x)
Answer: Let \( I = \int e^x(\sin x + \cos x) \, dx \).
We know the standard integral form \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \).
Here, let \( f(x) = \sin x \).
Then \( f'(x) = \cos x \).
So, the integral is of the form \( \int e^x [f(x) + f'(x)] \, dx \).
Thus, \( I = e^x \sin x + C \).
In simple words: This integral is a special type where \( e^x \) is multiplied by a function plus its derivative. In this case, \( \sin x \) is the function and \( \cos x \) is its derivative. So, the answer is just \( e^x \) times \( \sin x \), plus a constant.

Exam Tip: Always look for the pattern \( \int e^x[f(x)+f'(x)]dx \) as it offers a quick solution without needing integration by parts.

Question 17. \( \frac{x e^{x}}{(1+x)^{2}} \)
Answer: Let \( I = \int \frac{x e^x}{(1+x)^2} \, dx \).
We can rewrite the expression inside the integral to fit the form \( \int e^x [f(x) + f'(x)] \, dx \).
\( \frac{x}{(1+x)^2} = \frac{x+1-1}{(1+x)^2} = \frac{1+x}{(1+x)^2} - \frac{1}{(1+x)^2} = \frac{1}{1+x} - \frac{1}{(1+x)^2} \).
So, \( I = \int e^x [\frac{1}{1+x} - \frac{1}{(1+x)^2}] \, dx \).
Here, let \( f(x) = \frac{1}{1+x} \).
Then \( f'(x) = -\frac{1}{(1+x)^2} \).
Since it is in the form \( \int e^x [f(x) + f'(x)] \, dx \),
\( I = e^x f(x) + C \)
\( I = \frac{e^x}{1+x} + C \).
In simple words: We changed the fraction part to match a special rule for \( e^x \) integrals. By splitting \( \frac{x}{(1+x)^2} \) into two parts, we found one part was a function and the other was its derivative. This allowed us to quickly write down the answer as \( \frac{e^x}{1+x} \), plus the constant.

Exam Tip: For integrals involving \( e^x \) and rational functions, try to manipulate the rational function into the \( f(x) + f'(x) \) form using algebraic adjustments like adding and subtracting terms.

Question 18. \( e^{x}(\frac{1+\sin x}{1+\cos x}) \)
Answer: Let \( I = \int e^x \frac{1+\sin x}{1+\cos x} \, dx \).
We use the half-angle formulas: \( \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2} \) and \( 1+\cos x = 2 \cos^2 \frac{x}{2} \).
So, \( \frac{1+\sin x}{1+\cos x} = \frac{1+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \)
\( = \frac{1}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \)
\( = \frac{1}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2} \).
So, \( I = \int e^x (\tan \frac{x}{2} + \frac{1}{2} \sec^2 \frac{x}{2}) \, dx \).
This is in the form \( \int e^x [f(x) + f'(x)] \, dx \).
Here, let \( f(x) = \tan \frac{x}{2} \).
Then \( f'(x) = \frac{1}{2} \sec^2 \frac{x}{2} \).
Therefore, \( I = e^x \tan \frac{x}{2} + C \).
In simple words: To solve this, we used half-angle formulas to simplify the fraction inside the integral. This helped us change the expression into the special form \( e^x \) times a function plus its derivative. Once in that form, the answer is simply \( e^x \) times the function, which is \( \tan \frac{x}{2} \).

Exam Tip: Trigonometric identities, especially half-angle formulas, are crucial for simplifying integrands to fit common integration patterns like \( \int e^x[f(x)+f'(x)]dx \).

Question 19. \( e^{x}(\frac{1}{x} - \frac{1}{x^{2}}) \)
Answer: Let \( I = \int e^x (\frac{1}{x} - \frac{1}{x^2}) \, dx \).
This integral is in the standard form \( \int e^x [f(x) + f'(x)] \, dx \).
Here, let \( f(x) = \frac{1}{x} \).
Then \( f'(x) = -\frac{1}{x^2} \).
Since the integral matches this pattern, the solution is directly given by \( e^x f(x) + C \).
Therefore, \( I = e^x \cdot \frac{1}{x} + C = \frac{e^x}{x} + C \).
In simple words: This problem is a direct application of a special integration rule. We have \( e^x \) multiplied by a function, \( \frac{1}{x} \), and its derivative, \( -\frac{1}{x^2} \). When this pattern occurs, the answer is just \( e^x \) times the original function, plus the constant.

Exam Tip: Identify the \( f(x) \) and \( f'(x) \) components clearly in \( \int e^x[f(x)+f'(x)]dx \) type problems to quickly arrive at the solution \( e^x f(x) + C \).

Question 20. \( \frac{(x-3) e^{x}}{(x-1)^{3}} \)
Answer: Let \( I = \int \frac{(x-3) e^x}{(x-1)^3} \, dx \).
We can rewrite the fraction to fit the form \( f(x) + f'(x) \).
\( \frac{x-3}{(x-1)^3} = \frac{(x-1)-2}{(x-1)^3} = \frac{x-1}{(x-1)^3} - \frac{2}{(x-1)^3} = \frac{1}{(x-1)^2} - \frac{2}{(x-1)^3} \).
So, \( I = \int e^x [\frac{1}{(x-1)^2} - \frac{2}{(x-1)^3}] \, dx \).
This is in the form \( \int e^x [f(x) + f'(x)] \, dx \).
Here, let \( f(x) = \frac{1}{(x-1)^2} = (x-1)^{-2} \).
Then \( f'(x) = -2(x-1)^{-3} \cdot 1 = -\frac{2}{(x-1)^3} \).
Thus, \( I = e^x f(x) + C \)
\( I = \frac{e^x}{(x-1)^2} + C \).
In simple words: We changed the fraction part of the integral to match a special rule for \( e^x \). By rewriting \( \frac{x-3}{(x-1)^3} \) as two separate terms, we found one term was a function and the other was its derivative. This meant the answer was simply \( e^x \) times the function \( \frac{1}{(x-1)^2} \), plus the constant.

Exam Tip: Algebraic manipulation is key for integrals like this. Always attempt to express the rational function as a sum of a function and its derivative when \( e^x \) is present.

Question 21. e2xsinx
Answer: Let \( I = \int e^{2x} \sin x \, dx \).
Using integration by parts, with \( u = \sin x \) and \( dv = e^{2x} \, dx \).
So, \( du = \cos x \, dx \) and \( v = \frac{e^{2x}}{2} \).
\( I = \sin x \cdot \frac{e^{2x}}{2} - \int \frac{e^{2x}}{2} \cos x \, dx \)
\( I = \frac{e^{2x} \sin x}{2} - \frac{1}{2} \int e^{2x} \cos x \, dx \).
Now, integrate \( \int e^{2x} \cos x \, dx \) using integration by parts again, with \( u = \cos x \) and \( dv = e^{2x} \, dx \).
So, \( du = -\sin x \, dx \) and \( v = \frac{e^{2x}}{2} \).
\( \int e^{2x} \cos x \, dx = \cos x \cdot \frac{e^{2x}}{2} - \int \frac{e^{2x}}{2} (-\sin x) \, dx \)
\( = \frac{e^{2x} \cos x}{2} + \frac{1}{2} \int e^{2x} \sin x \, dx \).
Substitute this back into the equation for \( I \):
\( I = \frac{e^{2x} \sin x}{2} - \frac{1}{2} [\frac{e^{2x} \cos x}{2} + \frac{1}{2} \int e^{2x} \sin x \, dx] \)
\( I = \frac{e^{2x} \sin x}{2} - \frac{e^{2x} \cos x}{4} - \frac{1}{4} \int e^{2x} \sin x \, dx \).
Notice that \( \int e^{2x} \sin x \, dx \) is our original integral \( I \).
So, \( I = \frac{e^{2x} \sin x}{2} - \frac{e^{2x} \cos x}{4} - \frac{1}{4} I \).
Now, solve for \( I \):
\( I + \frac{1}{4} I = \frac{e^{2x} \sin x}{2} - \frac{e^{2x} \cos x}{4} \)
\( \frac{5}{4} I = \frac{2e^{2x} \sin x - e^{2x} \cos x}{4} \)
\( 5I = e^{2x} (2 \sin x - \cos x) \)
\( I = \frac{e^{2x}}{5} (2 \sin x - \cos x) + C \).
In simple words: This integral needed the integration by parts rule to be used two times. After applying the rule twice, we found that the original integral 'I' reappeared on the right side. We then rearranged the equation to solve for 'I', giving us the final expression.

Exam Tip: For integrals of the form \( \int e^{ax} \sin(bx) \, dx \) or \( \int e^{ax} \cos(bx) \, dx \), expect to apply integration by parts twice and then solve an algebraic equation for the integral.

Question 22. \( \sin^{-1}(\frac{2x}{1+x^{2}}) \)
Answer: Let \( I = \int \sin^{-1}(\frac{2x}{1+x^2}) \, dx \).
Let \( x = \tan t \). Then \( dx = \sec^2 t \, dt \).
The term \( \sin^{-1}(\frac{2x}{1+x^2}) \) becomes \( \sin^{-1}(\frac{2 \tan t}{1+\tan^2 t}) \).
We know that \( \frac{2 \tan t}{1+\tan^2 t} = \sin 2t \).
So, \( \sin^{-1}(\sin 2t) = 2t \).
Thus, \( I = \int 2t \sec^2 t \, dt \).
Using integration by parts, with \( u = 2t \) and \( dv = \sec^2 t \, dt \).
So, \( du = 2 \, dt \) and \( v = \tan t \).
\( I = 2t \tan t - \int (\tan t \cdot 2) \, dt \)
\( = 2t \tan t - 2 \int \tan t \, dt \)
\( = 2t \tan t - 2 \log|\cos t| + C \).
Now, substitute back \( t = \tan^{-1} x \).
Since \( x = \tan t \), we have \( \tan t = x \).
Also, \( \sec t = \sqrt{1+\tan^2 t} = \sqrt{1+x^2} \), so \( \cos t = \frac{1}{\sqrt{1+x^2}} \).
\( \log|\cos t| = \log|\frac{1}{\sqrt{1+x^2}}| = \log((1+x^2)^{-1/2}) = -\frac{1}{2} \log(1+x^2) \).
So, \( I = 2x \tan^{-1} x - 2 (-\frac{1}{2} \log(1+x^2)) + C \)
\( = 2x \tan^{-1} x + \log(1+x^2) + C \).
In simple words: We started by using a trigonometric substitution \( x = \tan t \), which simplified the inverse sine part to just \( 2t \). Then, we had to integrate \( 2t \sec^2 t \), which we did using integration by parts. Finally, we changed all the 't' terms back to 'x' to get our answer.

Exam Tip: Recognize the identity \( \sin^{-1}(\frac{2x}{1+x^2}) = 2 \tan^{-1}x \) (for \( |x| \le 1 \)). This direct substitution often simplifies the problem significantly.

Choose the Correct Answers in the Following Questions From 23 and 24:

Question 23. \( \int x^{2} e^{x^{3}} \, dx \) equals
(a) \( \frac{1}{3} e^{x^{3}} + C \)
(b) \( \frac{1}{3} e^{x^{2}} + C \)
(c) \( \frac{1}{2} e^{x^{3}} + C \)
(d) \( \frac{1}{2} e^{x^{2}} + C \)
Answer: (a) \( \frac{1}{3} e^{x^{3}} + C \)
In simple words: This integral can be solved using a simple substitution. If you let \( t = x^3 \), then \( dt = 3x^2 dx \). The integral then becomes \( \frac{1}{3} \int e^t dt \), which is \( \frac{1}{3} e^t + C \). Substituting back \( t = x^3 \) gives the answer.

Exam Tip: Always look for a suitable substitution, especially when the derivative of a part of the function is also present in the integrand.

Question 24. \( \int e^{x} \sec x(1 + \tan x) \, dx \) equals
(a) \( e^{x} \cos x + C \)
(b) \( e^{x} \sec x + C \)
(c) \( e^{x} \sin x + C \)
(d) \( e^{x} \tan x + C \)
Answer: (b) \( e^{x} \sec x + C \)
In simple words: This integral is a special type where \( e^x \) is multiplied by a function plus its derivative. If you expand the terms, you get \( e^x (\sec x + \sec x \tan x) \). Here, \( \sec x \) is the function, and \( \sec x \tan x \) is its derivative. So, the answer is \( e^x \sec x + C \).

Exam Tip: Recognize the pattern \( \int e^x[f(x)+f'(x)]dx \) by first distributing \( e^x \) and then identifying the function \( f(x) \) and its derivative \( f'(x) \).