CBSE Class 12 Mathematics HOTs Integration Set 02

Very Short Answer Type Questions

Question. Find \( \int \frac{2}{1 - \cos 2x} \, dx \)
Answer: Let \( I = \int \frac{2}{1 - \cos 2x} \, dx = \int \frac{2}{1 - (1 - 2\sin^2 x)} \, dx \)
\( = \int \frac{2}{2\sin^2 x} \, dx = \int \csc^2 x \, dx = -\cot x + C \)

Question. Evaluate \( \int \frac{\cos 2x + 2\sin^2 x}{\sin^2 x} \, dx \)
Answer: Let \( I = \int \frac{\cos 2x + 2\sin^2 x}{\sin^2 x} \, dx \)
\( = \int \frac{1 - 2\sin^2 x + 2\sin^2 x}{\sin^2 x} \, dx = \int \frac{1}{\sin^2 x} \, dx \)
\( = \int \csc^2 x \, dx = -\cot x + C \)

Question. Evaluate \( \int \frac{x^4 + x^2 + 1}{x^2 - x + 1} \, dx \)
Answer: Let \( I = \int \frac{x^4 + x^2 + 1}{x^2 - x + 1} \, dx = \int \frac{(x^2 - x + 1)(x^2 + x + 1)}{x^2 - x + 1} \, dx \)
\( = \int (x^2 + x + 1) \, dx = \int x^2 \, dx + \int x \, dx + \int 1 \, dx \)
\( = \frac{x^3}{3} + \frac{x^2}{2} + x + C \)

Question. Evaluate \( \int \sin x \left( \cot x + \frac{1}{\sin^3 x} \right) \, dx \)
Answer: Let \( I = \int \sin x \left( \cot x + \frac{1}{\sin^3 x} \right) \, dx \)
\( = \int \left( \sin x \times \cot x + \frac{\sin x}{\sin^3 x} \right) \, dx \)
\( = \int \left( \sin x \times \frac{\cos x}{\sin x} \right) \, dx + \int \frac{1}{\sin^2 x} \, dx \)
\( = \int \cos x \, dx + \int \csc^2 x \, dx \)
\( = \sin x + (-\cot x) + C \)
\( = \sin x - \cot x + C \)

Question. Evaluate \( \int \frac{x^4}{x^2 + 1} \, dx \)
Answer: Let \( I = \int \frac{x^4}{x^2 + 1} \, dx = \int \frac{x^4 - 1 + 1}{x^2 + 1} \, dx \)
\( = \int \frac{x^4 - 1}{x^2 + 1} \, dx + \int \frac{1}{x^2 + 1} \, dx \)
\( = \int \frac{(x^2 - 1)(x^2 + 1)}{x^2 + 1} \, dx + \int \frac{1}{x^2 + 1} \, dx \)      \( [\dots a^4 - b^4 = (a^2 - b^2)(a^2 + b^2)] \)
\( = \int (x^2 - 1) \, dx + \int \frac{1}{x^2 + 1} \, dx \)
\( = \int x^2 \, dx - \int dx + \int \frac{dx}{x^2 + 1} \)
\( = \frac{x^3}{3} - x + \tan^{-1} x + C \)      \( \left[ \because \int \frac{dx}{x^2 + 1} = \tan^{-1} x + C \right] \)

Question. Evaluate \( \int \frac{\cos 2x + 2\sin^2 x}{\cos^2 x} \, dx \)
Answer: Let \( I = \int \frac{\cos 2x + 2\sin^2 x}{\cos^2 x} \, dx \)
\( = \int \frac{1 - 2\sin^2 x + 2\sin^2 x}{\cos^2 x} \, dx \)
\( = \int \frac{1}{\cos^2 x} \, dx = \int \sec^2 x \, dx = \tan x + C \)

Question. Evaluate \( \int \frac{1}{e^x + 1} \, dx \)
Answer: Let \( I = \int \frac{1}{e^x + 1} \, dx = \int \frac{e^{-x}}{1 + e^{-x}} \, dx \)
Put \( 1 + e^{-x} = t \)
\( \Rightarrow -e^{-x} \, dx = dt \Rightarrow e^{-x} \, dx = -dt \)
\( \therefore I = -\int \frac{dt}{t} = -\log|t| + C \)
\( = -\log|1 + e^{-x}| + C \)

Question. Evaluate \( \int \sin^3 x \cos^5 x \, dx \)
Answer: Let \( I = \int \sin^3 x \cos^5 x \, dx \)
Here, the powers of both \( \sin x \) and \( \cos x \) are odd. So, either put \( \sin x = t \) or \( \cos x = t \).
Now, put \( \cos x = t \Rightarrow -\sin x \, dx = dt \Rightarrow dx = \frac{-dt}{\sin x} \)
\( \therefore I = \int \sin^3 x \cdot t^5 \left( \frac{-dt}{\sin x} \right) \)
\( = -\int \sin^2 x \cdot t^5 \, dt \)
\( = -\int (1 - \cos^2 x) t^5 \, dt \)
\( = -\int (1 - t^2) t^5 \, dt \)      \( [\because \cos x = t] \)
\( = -\int (t^5 - t^7) \, dt = \int (t^7 - t^5) \, dt \)
\( = \frac{t^8}{8} - \frac{t^6}{6} + C \)
\( = \frac{\cos^8 x}{8} - \frac{\cos^6 x}{6} + C \)      \( [\dots \cos x = t] \)

Question. Find \( \int \cos^3 x e^{\log \sin x} \, dx \)
Answer: Let \( I = \int \cos^3 x e^{\log \sin x} \, dx \)
\( = \int \cos^3 x \sin x \, dx \)      \( [\because e^{\log x} = x] \)
On putting \( \cos x = t \)
\( \Rightarrow -\sin x \, dx = dt \)
\( \Rightarrow \sin x \, dx = -dt \)
\( \therefore I = -\int t^3 \, dt = -\frac{t^4}{4} + C = -\frac{\cos^4 x}{4} + C \)

Question. Evaluate \( \int \tan^{-1} \sqrt{\frac{1 + \cos x}{1 - \cos x}} \, dx \)
Answer: Let \( I = \int \tan^{-1} \sqrt{\frac{1 + \cos x}{1 - \cos x}} \, dx = \int \tan^{-1} \sqrt{\frac{2\cos^2 \frac{x}{2}}{2\sin^2 \frac{x}{2}}} \, dx \)
     \( [\because \cos 2\theta = 2\cos^2 \theta - 1 = 1 - 2\sin^2 \theta] \)
\( = \int \tan^{-1} \sqrt{\cot^2 \frac{x}{2}} \, dx = \int \tan^{-1} \left( \cot \frac{x}{2} \right) \, dx \)
\( = \int \tan^{-1} \left[ \tan \left( \frac{\pi}{2} - \frac{x}{2} \right) \right] \, dx \)      \( \left[ \because \tan \left( \frac{\pi}{2} - \theta \right) = \cot \theta \right] \)
\( = \int \left( \frac{\pi}{2} - \frac{x}{2} \right) \, dx \)      \( [\because \tan^{-1}(\tan \theta) = \theta] \)
\( = \frac{\pi}{2} x - \frac{x^2}{2 \cdot 2} + C = \frac{\pi}{2} x - \frac{x^2}{4} + C \)

Question. Find \( \int \frac{dx}{\sqrt{9 - 25x^2}} \)
Answer: Let \( I = \int \frac{dx}{\sqrt{9 - 25x^2}} = \int \frac{dx}{\sqrt{25 \left( \frac{9}{25} - x^2 \right)}} \)
\( = \frac{1}{5} \int \frac{dx}{\sqrt{\left( \frac{3}{5} \right)^2 - x^2}} \)
\( = \frac{1}{5} \sin^{-1} \left( \frac{x}{3/5} \right) + C \)
\( = \frac{1}{5} \sin^{-1} \left( \frac{5x}{3} \right) + C \)

Short Answer Type Questions

Question. Evaluate \( \int \frac{x^3 - x^2 + x - 1}{x - 1} \, dx \)
Answer: Let \( I = \int \frac{x^3 - x^2 + x - 1}{x - 1} \, dx \)
\( = \int \frac{x^2(x - 1) + 1(x - 1)}{x - 1} \, dx = \int \frac{(x-1)(x^2 + 1)}{x-1} \, dx \)
\( = \int (x^2 + 1) \, dx = \frac{x^3}{3} + x + C \)

Question. Find \( \int \frac{3 - 5\sin x}{\cos^2 x} \, dx \)
Answer: Let \( I = \int \frac{3 - 5\sin x}{\cos^2 x} \, dx \)
\( = \int \left( \frac{3}{\cos^2 x} - \frac{5\sin x}{\cos^2 x} \right) \, dx \)
\( = 3\int \sec^2 x \, dx - 5\int \sec x \tan x \, dx \)
\( = 3\tan x - 5\sec x + C \)

Question. Find \( \int \frac{dx}{\cos x \sqrt{\cos 2x}} \)
Answer: Let \( I = \int \frac{dx}{\cos x \sqrt{\cos 2x}} \)
\( = \int \frac{dx}{\cos x \sqrt{1 - 2\sin^2 x}} \)      \( [\because \cos 2\theta = 1 - 2\sin^2 \theta] \)
\( = \int \frac{\sec^2 x \, dx}{\sqrt{\frac{1}{\cos^2 x} - \frac{2\sin^2 x}{\cos^2 x}}} \)      [dividing numerator and denominator by \( \cos^2 x \)]
\( = \int \frac{\sec^2 x \, dx}{\sqrt{\sec^2 x - 2\tan^2 x}} \)
\( = \int \frac{\sec^2 x \, dx}{\sqrt{1 + \tan^2 x - 2\tan^2 x}} \)      \( [\because 1 + \tan^2 \theta = \sec^2 \theta] \)
\( = \int \frac{\sec^2 x \, dx}{\sqrt{1 - \tan^2 x}} \)
Again, let \( \tan x = t \Rightarrow \sec^2 x \, dx = dt \)
\( \dots I = \int \frac{dt}{\sqrt{1 - t^2}} \)
\( = \sin^{-1} t + C \)      \( \left[ \because \int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1} x + C \right] \)
\( = \sin^{-1}(\tan x) + C \)      \( [\because t = \tan x] \)

Question. Evaluate \( \int \frac{x^3}{(x+2)^4} \, dx \)
Answer: Let \( I = \int \frac{x^3}{(x+2)^4} \, dx \)
Put \( x + 2 = t \Rightarrow dx = dt \) and \( x = t - 2 \)
\( \dots I = \int \frac{(t-2)^3}{t^4} \, dt \)
\( = \int \frac{t^3 - 6t^2 + 12t - 8}{t^4} \, dt \)      \( [\because (a-b)^3 = a^3 - b^3 - 3a^2b + 3ab^2] \)
\( = \int \left( \frac{1}{t} - \frac{6}{t^2} + \frac{12}{t^3} - \frac{8}{t^4} \right) \, dt \)
\( = \log t + \frac{6}{t} - \frac{6}{t^2} + \frac{8}{3t^3} + C \)
\( = \log|x+2| + \frac{6}{x+2} - \frac{6}{(x+2)^2} + \frac{8}{3(x+2)^3} + C \)

Question. Evaluate \( \int \frac{3x+1}{(3x^2 + 2x + 1)^3} \, dx \)
Answer: Let \( I = \int \frac{3x+1}{(3x^2+2x+1)^3} \, dx \)
Put \( 3x^2 + 2x + 1 = t \)
\( \Rightarrow (6x+2) \, dx = dt \)
\( \Rightarrow (3x+1) \, dx = \frac{1}{2} \, dt \)
\( \dots I = \frac{1}{2} \int \frac{dt}{t^3} = \frac{1}{2} \int t^{-3} \, dt \)
\( = \frac{1}{2} \frac{t^{-2}}{(-2)} + C = \frac{-1}{4t^2} + C \)
\( = -\frac{1}{4(3x^2 + 2x + 1)^2} + C \)

Question. Evaluate \( \int \cos 6x \sqrt{1 + \sin 6x} \, dx \)
Answer: Let \( I = \int \cos 6x \sqrt{1 + \sin 6x} \, dx \)
Put \( 1 + \sin 6x = t \Rightarrow 6\cos 6x \, dx = dt \Rightarrow \cos 6x \, dx = \frac{dt}{6} \)
\( \dots I = \frac{1}{6} \int \sqrt{t} \, dt = \frac{1}{6} \left( \frac{t^{1/2 + 1}}{\frac{1}{2} + 1} \right) + C \)
\( = \frac{1}{6} \frac{t^{3/2}}{3/2} + C \)
\( = \frac{1}{6} \times \frac{2}{3} t^{3/2} + C = \frac{1}{9} t^{3/2} + C \)
\( = \frac{1}{9}(1 + \sin 6x)^{3/2} + C \)

Question. Evaluate \( \int \sin 5x \cos 3x \, dx \)
Answer: Let \( I = \int \sin 5x \cos 3x \, dx \)
\( = \frac{1}{2} \int 2\sin 5x \cos 3x \, dx \)      [multiply numerator and denominator by 2]
\( = \frac{1}{2} \int [\sin(5x+3x) + \sin(5x-3x)] \, dx \)
\( = \frac{1}{2} \int (\sin 8x + \sin 2x) \, dx \)
\( = -\frac{1}{2} \left[ \frac{\cos 8x}{8} + \frac{\cos 2x}{2} \right] + C \)
\( = -\frac{1}{16}\cos 8x - \frac{1}{4}\cos 2x + C \)

Question. Evaluate \( \int \frac{dx}{\sin^2 x \cos^2 x} \)
Answer: Let \( I = \int \frac{dx}{\sin^2 x \cos^2 x} \)
On dividing the numerator and denominator by \( \cos^4 x \), we get
\( I = \int \frac{\sec^2 x \cdot \sec^2 x}{\tan^2 x} \, dx \)
\( \Rightarrow I = \int \frac{(1 + \tan^2 x) \sec^2 x}{\tan^2 x} \, dx \)
Put \( \tan x = t \Rightarrow \sec^2 x \, dx = dt \)
\( \dots I = \int \frac{1+t^2}{t^2} \, dt \)
\( = \int \left( \frac{1}{t^2} + 1 \right) \, dt = \int t^{-2} \, dt + \int 1 \, dt \)
\( = -\frac{1}{t} + t + C \)
\( \Rightarrow I = \tan x - \cot x + C \)      \( [\because t = \tan x] \)

Question. Evaluate \( \int \cos^{-1} (\sin x) \, dx \)
Answer: Let \( I = \int \cos^{-1}(\sin x) \, dx \)
\( = \int \cos^{-1} \left[ \cos \left( \frac{\pi}{2} - x \right) \right] \, dx \)
\( = \int \left( \frac{\pi}{2} - x \right) \, dx \)      \( [\because \cos^{-1}(\cos \theta) = \theta] \)
\( = \frac{\pi}{2} \int dx - \int x \, dx \)
\( = \frac{\pi}{2} x - \frac{x^2}{2} + C \)

Question. Evaluate \( \int \frac{1}{\sqrt{x^2 - 4x}} \, dx \)
Answer: Let \( I = \int \frac{dx}{\sqrt{x^2 - 4x}} \)
\( = \int \frac{dx}{\sqrt{x^2 - 2 \cdot 2 \cdot x + 2^2 - 2^2}} = \int \frac{dx}{\sqrt{(x-2)^2 - 2^2}} \)
\( = \log\left| (x-2) + \sqrt{(x-2)^2 - 2^2} \right| + C \)      \( \left[ \because \int \frac{dx}{\sqrt{x^2 - a^2}} = \log \left| x + \sqrt{x^2 - a^2} \right| + C \right] \)
\( = \log\left| (x-2) + \sqrt{x^2 - 4x} \right| + C \)

Question. Find the value of \( \int \frac{\tan^2 x \cdot \sec^2 x}{1 - \tan^6 x} \, dx \)
Answer: Let \( I = \int \frac{\tan^2 x \cdot \sec^2 x}{1 - \tan^6 x} \, dx \)
Let \( \tan^3 x = t \)
\( \Rightarrow 3\tan^2 x \sec^2 x \, dx = dt \Rightarrow \tan^2 x \sec^2 x \, dx = \frac{1}{3} \, dt \)
\( \dots I = \frac{1}{3} \int \frac{dt}{1-t^2} = \frac{1}{3} \cdot \frac{1}{2} \log\left| \frac{1+t}{1-t} \right| + C \)      \( \left[ \times \int \frac{dx}{a^2-x^2} = \frac{1}{2a} \log\left| \frac{a+x}{a-x} \right| + C \right] \)
\( = \frac{1}{6} \log\left| \frac{1+t}{1-t} \right| + C \)
Now, put the value of \( t \), we get:
\( I = \frac{1}{6} \log\left| \frac{1+\tan^3 x}{1-\tan^3 x} \right| + C \)      \( [\because \tan^3 x = t] \)

Question. Find the value of \( \int \frac{\cos x}{(1 + \sin x)(2 + \sin x)} \, dx \)
Answer: Let \( I = \int \frac{\cos x}{(1 + \sin x)(2 + \sin x)} \, dx \)
Put \( \sin x = t \Rightarrow \cos x \, dx = dt \)
\( \dots I = \int \frac{dt}{(1+t)(2+t)} \)
Let \( \frac{1}{(1+t)(2+t)} = \frac{A}{1+t} + \frac{B}{2+t} \)      ...(i)
\( \dots \frac{1}{(1+t)(2+t)} = \frac{A(2+t) + B(1+t)}{(1+t)(2+t)} \)
\( \Rightarrow 1 = 2A + tA + B + tB \Rightarrow 1 = (2A + B) + t(A + B) \)
On comparing the coefficients of \( t \) and constant terms from both sides, we get:
\( 2A + B = 1 \) and \( A + B = 0 \)
\( \Rightarrow A = 1 \) and \( B = -1 \)
\( \dots I = \int \left( \frac{1}{1+t} - \frac{1}{2+t} \right) \, dt \)
\( = \int \frac{1}{1+t} \, dt - \int \frac{1}{2+t} \, dt \)
\( = \log|1+t| - \log|2+t| + C \)
\( = \log\left| \frac{1+t}{2+t} \right| + C = \log\left| \frac{1+\sin x}{2+\sin x} \right| + C \)      \( [\because t = \sin x] \)

Question. Find \( \int \frac{4}{(x-2)(x^2 + 4)} \, dx \)
Answer: Let \( I = \int \frac{4}{(x-2)(x^2 + 4)} \, dx \)
Again, let \( \frac{4}{(x-2)(x^2 + 4)} = \frac{A}{x-2} + \frac{Bx + C}{x^2 + 4} \)
\( \Rightarrow 4 = A(x^2 + 4) + (Bx + C)(x - 2) \)
\( \Rightarrow 4 = x^2(A + B) + x(-2B + C) + 4A - 2C \)
On equating the coefficients of \( x^2, x \) and constant terms from both sides, we get:
\( A + B = 0 \)      ...(i)
\( -2B + C = 0 \)      ...(ii)
\( 4A - 2C = 4 \)      ...(iii)
On solving Eqs. (i), (ii) and (iii), we get:
\( A = \frac{1}{2}, B = -\frac{1}{2} \) and \( C = -1 \)
\( \dots I = \int \frac{4}{(x-2)(x^2 + 4)} \, dx \)
\( = \int \frac{1/2}{x-2} \, dx + \int \frac{-\frac{1}{2}x - 1}{x^2 + 4} \, dx \)
\( = \frac{1}{2} \int \frac{dx}{x-2} - \int \frac{x+2}{2(x^2 + 4)} \, dx \)
\( = \frac{1}{2} \log|x-2| - \frac{1}{4} \log|x^2 + 4| - \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + C \)

Question. Evaluate \( \int x^3 \log 2x \, dx \)
Answer: Let \( I = \int x^3 \log 2x \, dx \)
Using integration by parts:
\( I = \log 2x \int x^3 \, dx - \int \left[ \frac{d}{dx}(\log 2x) \int x^3 \, dx \right] \, dx \)
\( = \log 2x \cdot \frac{x^4}{4} - \int \frac{1}{2x} \cdot 2 \cdot \frac{x^4}{4} \, dx \)
\( = \frac{x^4}{4} \log 2x - \frac{1}{4} \int x^3 \, dx \)
\( = \frac{x^4}{4} \log 2x - \frac{1}{16} x^4 + C \)

Long Answer Type Questions

Question. Evaluate \(\int \cos 2x \cos 4x \cos 6x dx\)
Answer: Let \(I = \int \cos 2x \cos 4x \cos 6x \, dx\)
\(= \frac{1}{2} \int (2 \cos 4x \cos 2x) \cos 6x \, dx\)
Using \(2 \cos A \cos B = \cos(A+B) + \cos(A-B)\):
\(I = \frac{1}{2} \int [\cos(4x+2x) + \cos(4x-2x)] \cos 6x \, dx\)
\(= \frac{1}{2} \int (\cos 6x + \cos 2x) \cos 6x \, dx\)
\(= \frac{1}{2} \int (\cos^2 6x + \cos 6x \cos 2x) \, dx\)
\(= \frac{1}{4} \int (2\cos^2 6x + 2\cos 6x \cos 2x) \, dx\)
Using \(2\cos^2 \theta = 1 + \cos 2\theta\) and \(2\cos A \cos B = \cos(A+B) + \cos(A-B)\):
\(I = \frac{1}{4} \int (1 + \cos 12x + \cos 8x + \cos 4x) \, dx\)
\(= \frac{1}{4} \left[ x + \frac{\sin 12x}{12} + \frac{\sin 8x}{8} + \frac{\sin 4x}{4} \right] + C\)
\(= \frac{x}{4} + \frac{\sin 12x}{48} + \frac{\sin 8x}{32} + \frac{\sin 4x}{16} + C\)

Question. Evaluate \(\int \frac{\sin^6 x + \cos^6 x}{\sin^2 x \cos^2 x} dx\)
Answer: Let \(I = \int \frac{\sin^6 x + \cos^6 x}{\sin^2 x \cos^2 x} dx\)
Since \(\sin^6 x + \cos^6 x = (\sin^2 x)^3 + (\cos^2 x)^3 = (\sin^2 x + \cos^2 x)^3 - 3 \sin^2 x \cos^2 x (\sin^2 x + \cos^2 x)\)
Using \(\sin^2 x + \cos^2 x = 1\), we have:
\(\sin^6 x + \cos^6 x = 1 - 3 \sin^2 x \cos^2 x\)
Therefore, \(I = \int \frac{1 - 3 \sin^2 x \cos^2 x}{\sin^2 x \cos^2 x} dx\)
\(= \int \frac{1}{\sin^2 x \cos^2 x} dx - 3 \int 1 \, dx\)
\(= \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} dx - 3x\)
\(= \int (\sec^2 x + \csc^2 x) dx - 3x\)
\(= \tan x - \cot x - 3x + C\)

Question. Evaluate \(\int \frac{dx}{a^2 \sin^2 x + b^2 \cos^2 x}\)
Answer: Let \(I = \int \frac{dx}{a^2 \sin^2 x + b^2 \cos^2 x}\).
Dividing the numerator and the denominator by \(\cos^2 x\), we get:
\(I = \int \frac{\sec^2 x}{a^2 \tan^2 x + b^2} dx = \int \frac{\sec^2 x}{(a \tan x)^2 + b^2} dx\)
Put \(a \tan x = t \Rightarrow a \sec^2 x \, dx = dt \Rightarrow \sec^2 x \, dx = \frac{dt}{a}\).
Now, \(I = \frac{1}{a} \int \frac{dt}{t^2 + b^2} = \frac{1}{ab} \tan^{-1}\left(\frac{t}{b}\right) + C = \frac{1}{ab} \tan^{-1}\left(\frac{a \tan x}{b}\right) + C\)

Question. Find \(\int \frac{x^2+x+1}{(x+2)(x^2+1)} dx\)
Answer: Let \(I = \int \frac{x^2+x+1}{(x+2)(x^2+1)} dx\).
Using partial fractions:
\(\frac{x^2+x+1}{(x+2)(x^2+1)} = \frac{A}{x+2} + \frac{Bx+C}{x^2+1}\)
\(\Rightarrow x^2 + x + 1 = A(x^2+1) + (Bx+C)(x+2)\)
Putting \(x = -2\) in this identity:
\(4 - 2 + 1 = A(5) \Rightarrow 5A = 3 \Rightarrow A = \frac{3}{5}\).
Putting \(x = 0\):
\(1 = A + 2C \Rightarrow 1 = \frac{3}{5} + 2C \Rightarrow 2C = \frac{2}{5} \Rightarrow C = \frac{1}{5}\).
Putting \(x = 1\):
\(3 = 2A + 3(B+C) \Rightarrow 3 = 2\left(\frac{3}{5}\right) + 3\left(B + \frac{1}{5}\right)\)
\(\Rightarrow 3 = \frac{6}{5} + 3B + \frac{3}{5} = \frac{9}{5} + 3B \Rightarrow 3B = \frac{6}{5} \Rightarrow B = \frac{2}{5}\).
Thus, \(\frac{x^2+x+1}{(x+2)(x^2+1)} = \frac{3}{5(x+2)} + \frac{2x+1}{5(x^2+1)}\).
Therefore, \(I = \frac{3}{5} \int \frac{dx}{x+2} + \frac{1}{5} \int \frac{2x}{x^2+1} dx + \frac{1}{5} \int \frac{dx}{x^2+1}\)
\(= \frac{3}{5} \log|x+2| + \frac{1}{5} \log(x^2+1) + \frac{1}{5} \tan^{-1} x + C\)

Question. Evaluate \(\int \frac{x}{(x^2+1)(x-1)} dx\)
Answer: Let \(I = \int \frac{x}{(x^2+1)(x-1)} dx\).
Using partial fractions:
\(\frac{x}{(x^2+1)(x-1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}\)
\(\Rightarrow x = A(x^2+1) + (Bx+C)(x-1)\)
Putting \(x = 1\):
\(1 = 2A \Rightarrow A = \frac{1}{2}\).
Equating the coefficients of \(x^2\) and constant terms:
\(0 = A + B \Rightarrow B = -A = -\frac{1}{2}\).
\(0 = A - C \Rightarrow C = A = \frac{1}{2}\).
Thus, \(\frac{x}{(x^2+1)(x-1)} = \frac{1}{2(x-1)} + \frac{-x+1}{2(x^2+1)}\).
Therefore, \(I = \frac{1}{2} \int \frac{dx}{x-1} - \frac{1}{4} \int \frac{2x}{x^2+1} dx + \frac{1}{2} \int \frac{dx}{x^2+1}\)
\(= \frac{1}{2} \log|x-1| - \frac{1}{4} \log(x^2+1) + \frac{1}{2} \tan^{-1} x + C\)

Question. Find \(\int \frac{\sec x}{1 + \csc x} dx\)
Answer: Let \(I = \int \frac{\sec x}{1 + \csc x} dx = \int \frac{\frac{1}{\cos x}}{1 + \frac{1}{\sin x}} dx = \int \frac{\sin x}{\cos x (1 + \sin x)} dx\)
Multiplying the numerator and the denominator by \(\cos x\):
\(I = \int \frac{\sin x \cos x}{\cos^2 x (1+\sin x)} dx = \int \frac{\sin x \cos x}{(1-\sin^2 x)(1+\sin x)} dx = \int \frac{\sin x \cos x}{(1-\sin x)(1+\sin x)^2} dx\)
Put \(\sin x = t \Rightarrow \cos x \, dx = dt\).
So, \(I = \int \frac{t}{(1-t)(1+t)^2} dt\).
Using partial fractions:
\(\frac{t}{(1-t)(1+t)^2} = \frac{A}{1+t} + \frac{B}{(1+t)^2} + \frac{C}{1-t}\)
\(\Rightarrow t = A(1+t)(1-t) + B(1-t) + C(1+t)^2\)
Putting \(t = -1\):
\(-1 = 2B \Rightarrow B = -\frac{1}{2}\).
Putting \(t = 1\):
\(1 = 4C \Rightarrow C = \frac{1}{4}\).
Putting \(t = 0\):
\(0 = A + B + C \Rightarrow A = -B - C = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}\).
Thus, \(I = \frac{1}{4} \int \frac{dt}{1+t} - \frac{1}{2} \int \frac{dt}{(1+t)^2} + \frac{1}{4} \int \frac{dt}{1-t}\)
\(= \frac{1}{4} \log|1+t| + \frac{1}{2(1+t)} - \frac{1}{4} \log|1-t| + C\)
Replacing \(t = \sin x\):
\(I = \frac{1}{4} \log|1+\sin x| + \frac{1}{2(1+\sin x)} - \frac{1}{4} \log|1-\sin x| + C\)

Question. Evaluate \(\int e^{ax} \sin bx \, dx\)
Answer: Let \(I = \int e^{ax} \sin bx \, dx\) — (i)
Integrating by parts taking \(\sin bx\) as the first function:
\(I = \sin bx \left( \frac{e^{ax}}{a} \right) - \int b \cos bx \left( \frac{e^{ax}}{a} \right) dx\)
\(I = \frac{e^{ax} \sin bx}{a} - \frac{b}{a} \int e^{ax} \cos bx \, dx\)
Integrating by parts again:
\(\int e^{ax} \cos bx \, dx = \cos bx \left( \frac{e^{ax}}{a} \right) - \int -b \sin bx \left( \frac{e^{ax}}{a} \right) dx = \frac{e^{ax} \cos bx}{a} + \frac{b}{a} I\)
Substituting this back into the expression for \(I\):
\(I = \frac{e^{ax} \sin bx}{a} - \frac{b}{a} \left[ \frac{e^{ax} \cos bx}{a} + \frac{b}{a} I \right]\)
\(I = \frac{e^{ax} \sin bx}{a} - \frac{b e^{ax} \cos bx}{a^2} - \frac{b^2}{a^2} I\)
\(I \left(1 + \frac{b^2}{a^2}\right) = \frac{e^{ax}}{a^2} [a\sin bx - b\cos bx]\)
\(I \left( \frac{a^2+b^2}{a^2} \right) = \frac{e^{ax}}{a^2} [a\sin bx - b\cos bx]\)
\(I = \frac{e^{ax}}{a^2+b^2} [a\sin bx - b\cos bx] + C\)