CBSE Class 12 Mathematics HOTs Integration Set 03

Very Short Answer Type Questions

Question. Find \( \int \frac{dx}{5 - 8x - x^2} \)
Answer: Let \( I = \int \frac{dx}{5 - 8x - x^2} \)
\( = \int \frac{dx}{5 - 2 \cdot 4 \cdot x - x^2 - (4)^2 + (4)^2} \)
\( = \int \frac{dx}{5 + 16 - [x^2 + (4)^2 + 2 \cdot 4 \cdot x]} \)
\( = \int \frac{dx}{21 - (x + 4)^2} = \int \frac{dx}{(\sqrt{21})^2 - (x+4)^2} \)
\( = \frac{1}{2\sqrt{21}} \log \left| \frac{\sqrt{21} + x + 4}{\sqrt{21} - x - 4} \right| + C \)      \( \left[ \because \int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C \right] \)

Question. Find \( \int \frac{e^x}{\sqrt{9 - e^{2x}}} \, dx \)
Answer: Let \( I = \int \frac{e^x}{\sqrt{9 - e^{2x}}} \, dx = \int \frac{e^x}{\sqrt{(3)^2 - (e^x)^2}} \, dx \)
Now, put \( e^x = t \Rightarrow e^x \, dx = dt \)
\( \therefore I = \int \frac{dt}{\sqrt{(3)^2 - t^2}} = \sin^{-1} \frac{t}{3} + C \)      \( \left[ \times \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \left( \frac{x}{a} \right) \right] \)
\( = \sin^{-1} \left( \frac{e^x}{3} \right) + C \)

Question. Evaluate \( \int x \sin 3x \, dx \)
Answer: Let \( I = \int x \sin 3x \, dx \)
Using integration by parts:
\( = x \int \sin 3x \, dx - \int \left[ \frac{d}{dx}(x) \times \int \sin 3x \, dx \right] \, dx \)
\( = x \left( -\frac{\cos 3x}{3} \right) - \int 1 \cdot \left( -\frac{\cos 3x}{3} \right) \, dx \)
\( = -\frac{1}{3} x \cos 3x + \frac{1}{3} \int \cos 3x \, dx \)
\( = -\frac{1}{3} x \cos 3x + \frac{1}{9} \sin 3x + C \)

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

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

Question. Find \( \int e^x \left( \frac{2 + \sin 2x}{1 + \cos 2x} \right) \, dx \)
Answer: Let \( I = \int e^x \left( \frac{2 + \sin 2x}{1 + \cos 2x} \right) \, dx \)
\( = \int e^x \left( \frac{2 + 2\sin x \cos x}{2\cos^2 x} \right) \, dx \)
\( = \int e^x (\sec^2 x + \tan x) \, dx \)
\( = \int e^x \sec^2 x \, dx + \int e^x \tan x \, dx \)
Using integration by parts on the second term:
\( = \int e^x \sec^2 x \, dx + \tan x (e^x) - \int e^x \sec^2 x \, dx \)
\( = e^x \tan x + C \)

Question. Evaluate \( \int \sin^{-1} 2x \, dx \)
Answer: Let \( I = \int \sin^{-1} 2x \, dx = \int 1 \cdot \sin^{-1} 2x \, dx \)
Using integration by parts:
\( = \sin^{-1} 2x \cdot x - \int x \cdot \frac{1}{\sqrt{1 - 4x^2}} (2) \, dx \)
\( = x \sin^{-1} 2x - \int \frac{2x}{\sqrt{1 - 4x^2}} \, dx \)
\( = x \sin^{-1} 2x + \frac{1}{4} \int \frac{-8x}{\sqrt{1 - 4x^2}} \, dx \)
Let \( I_1 = \int \frac{-8x}{\sqrt{1 - 4x^2}} \, dx \)
Let \( 1 - 4x^2 = t \Rightarrow -8x \, dx = dt \)
Then, \( I_1 = \int \frac{dt}{\sqrt{t}} = 2\sqrt{t} = 2\sqrt{1 - 4x^2} \)
\( \therefore I = x \sin^{-1} 2x + \frac{1}{4} \left( 2\sqrt{1 - 4x^2} \right) + C \)
\( = x \sin^{-1} 2x + \frac{1}{2}\sqrt{1 - 4x^2} + C \)

Question. Evaluate \( \int_{0}^{\pi/2} \sin^2 x \, dx \)
Answer: Let \( I = \int_{0}^{\pi/2} \sin^2 x \, dx = \int_{0}^{\pi/2} \left( \frac{1 - \cos 2x}{2} \right) \, dx \)
\( = \frac{1}{2} \left[ \int_{0}^{\pi/2} 1 \, dx - \int_{0}^{\pi/2} \cos 2x \, dx \right] \)
\( = \frac{1}{2} \left[ x - \frac{\sin 2x}{2} \right]_{0}^{\pi/2} = \frac{1}{2} \left[ \frac{\pi}{2} - \frac{\sin \pi}{2} \right] - \frac{1}{2} [0 - 0] \)
\( = \frac{1}{2} \left[ \frac{\pi}{2} - 0 \right] - 0 = \frac{\pi}{4} \)

Question. Evaluate \( \int_{0}^{3} \frac{dx}{9 + x^2} \)
Answer: Let \( I = \int_{0}^{3} \frac{dx}{9 + x^2} = \int_{0}^{3} \frac{dx}{x^2 + (3)^2} \)
\( \Rightarrow I = \left[ \frac{1}{3} \tan^{-1} \frac{x}{3} \right]_{0}^{3} \)      \( \left[ \because \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} \right] \)
\( \Rightarrow I = \frac{1}{3} \left[ \tan^{-1}(1) - \tan^{-1}(0) \right] \)
\( \Rightarrow I = \frac{1}{3} \left[ \frac{\pi}{4} - 0 \right] = \frac{\pi}{12} \)

Question. Evaluate \( \int_{0}^{1} x e^{x^2} \, dx \)
Answer: Let \( I = \int_{0}^{1} x e^{x^2} \, dx \)
Now, put \( x^2 = t \Rightarrow 2x \, dx = dt \)
Upper limit: When \( x = 1 \), then \( t = 1^2 = 1 \)
Lower limit: When \( x = 0 \), then \( t = 0 \)
\( \therefore I = \frac{1}{2} \int_{0}^{1} e^t \, dt = \frac{1}{2} [e^t]_{0}^{1} = \frac{1}{2}(e^1 - e^0) = \frac{1}{2}(e - 1) \)

Question. Evaluate \( \int_{-1}^{2} \frac{|x|}{x} \, dx \)
Answer: Let \( I = \int_{-1}^{2} \frac{|x|}{x} \, dx \)
\( = \int_{-1}^{0} \frac{|x|}{x} \, dx + \int_{0}^{2} \frac{|x|}{x} \, dx \)
\( = \int_{-1}^{0} \frac{-x}{x} \, dx + \int_{0}^{2} \frac{x}{x} \, dx \)      \( \left[ \because |x| = \begin{cases} -x, & x < 0 \\ x, & x \ge 0 \end{cases} \right] \)
\( = -\int_{-1}^{0} 1 \, dx + \int_{0}^{2} 1 \, dx = -[x]_{-1}^{0} + [x]_{0}^{2} \)
\( = -[0 - (-1)] + [2 - 0] = -1 + 2 = 1 \)

Short Answer Type Questions

Question. Evaluate \( \int_{0}^{\pi/2} \frac{x}{\cos x + \sin x} \, dx \)
Answer: Let \( I = \int_{0}^{\pi/2} \frac{x}{\cos x + \sin x} \, dx \)      ...(i)
\( \Rightarrow I = \int_{0}^{\pi/2} \frac{\frac{\pi}{2} - x}{\cos\left(\frac{\pi}{2} - x\right) + \sin\left(\frac{\pi}{2} - x\right)} \, dx \)
\( \Rightarrow I = \int_{0}^{\pi/2} \frac{\frac{\pi}{2} - x}{\sin x + \cos x} \, dx \)      ...(ii)
On adding Eqs. (i) and (ii), we get:
\( 2I = \int_{0}^{\pi/2} \frac{\frac{\pi}{2} - x + x}{\sin x + \cos x} \, dx \)
\( \Rightarrow 2I = \frac{\pi}{2} \int_{0}^{\pi/2} \frac{1}{\sin x + \cos x} \, dx \)
\( \Rightarrow 2I = \frac{\pi}{2} \int_{0}^{\pi/2} \frac{1}{\sin x + \cos x} \, dx \)
\( \Rightarrow I = \frac{\pi}{4} \int_{0}^{\pi/2} \frac{1}{\sin x + \cos x} \, dx \)

Question. Evaluate \( \int_{0}^{2} \frac{1}{\sqrt{x} + \sqrt{1+x}} \, dx \)
Answer: Let \( I = \int_{0}^{2} \frac{1}{\sqrt{x} + \sqrt{1+x}} \, dx \)
\( = \int_{0}^{2} \frac{\sqrt{1+x} - \sqrt{x}}{(\sqrt{1+x} + \sqrt{x})(\sqrt{1+x} - \sqrt{x})} \, dx \)
\( = \int_{0}^{2} (\sqrt{1+x} - \sqrt{x}) \, dx \)
\( = \left[ \frac{2}{3}(1+x)^{3/2} - \frac{2}{3}x^{3/2} \right]_{0}^{2} \)
\( = \frac{2}{3} (3^{3/2} - 1) - \frac{2}{3}(2)^{3/2} \)
\( = \frac{2}{3} \times 3\sqrt{3} - \frac{2}{3} - \frac{2}{3} \times 2\sqrt{2} \)
\( = 2\sqrt{3} - \frac{2}{3} - \frac{4}{3}\sqrt{2} \)

Question. Evaluate \( \int_{1}^{4} \frac{1}{\sqrt{2x+1} - \sqrt{2x-1}} \, dx \)
Answer: Let \( I = \int_{1}^{4} \frac{1}{\sqrt{2x+1} - \sqrt{2x-1}} \, dx \)
\( = \int_{1}^{4} \frac{\sqrt{2x+1} + \sqrt{2x-1}}{(\sqrt{2x+1} - \sqrt{2x-1})(\sqrt{2x+1} + \sqrt{2x-1})} \, dx \)
\( = \int_{1}^{4} \frac{\sqrt{2x+1} + \sqrt{2x-1}}{(2x+1) - (2x-1)} \, dx \)
\( = \frac{1}{2} \int_{1}^{4} \left[ (2x+1)^{1/2} + (2x-1)^{1/2} \right] \, dx \)
\( = \frac{1}{2} \left[ \int_{1}^{4} (2x+1)^{1/2} \, dx + \int_{1}^{4} (2x-1)^{1/2} \, dx \right] \)
\( = \frac{1}{2} \left[ \frac{1}{2} \frac{(2x+1)^{3/2}}{3/2} + \frac{1}{2} \frac{(2x-1)^{3/2}}{3/2} \right]_{1}^{4} \)
\( = \frac{1}{6} \left[ (2x+1)^{3/2} + (2x-1)^{3/2} \right]_{1}^{4} \)
\( = \frac{1}{6} \left[ \left( 9^{3/2} - 3^{3/2} \right) + \left( 7^{3/2} - 1^{3/2} \right) \right] \)
\( = \frac{1}{6} [27 - 3^{3/2} + 7^{3/2} - 1] \)
\( = \frac{1}{6}(26 - 3^{3/2} + 7^{3/2}) \)

Question. Evaluate \( \int_{0}^{\pi/4} \sqrt{1 - \sin 2x} \, dx \)
Answer: Let \( I = \int_{0}^{\pi/4} \sqrt{1 - \sin 2x} \, dx \)
\( = \int_{0}^{\pi/4} \sqrt{\cos^2 x + \sin^2 x - 2\sin x \cos x} \, dx \)
\( = \int_{0}^{\pi/4} \sqrt{(\cos x - \sin x)^2} \, dx = \int_{0}^{\pi/4} |\cos x - \sin x| \, dx \)
\( = \int_{0}^{\pi/4} (\cos x - \sin x) \, dx \)
\( = [\sin x + \cos x]_{0}^{\pi/4} \)
\( = \left( \sin \frac{\pi}{4} + \cos \frac{\pi}{4} \right) - (\sin 0 + \cos 0) \)
\( = \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) - (0 + 1) = \sqrt{2} - 1 \)

Question. Evaluate \(\int_0^1 \frac{2x+3}{5x^2+1} dx\)
Answer: Let \(I = \int_0^1 \frac{2x+3}{5x^2+1} dx = \int_0^1 \frac{2x}{5x^2+1} dx + \int_0^1 \frac{3}{5x^2+1} dx\)
\(= \frac{1}{5} \int_0^1 \frac{10x}{5x^2+1} dx + \frac{3}{5} \int_0^1 \frac{1}{x^2 + \left(\frac{1}{\sqrt{5}}\right)^2} dx\)
\(= \frac{1}{5} \left[ \log|5x^2+1| \right]_0^1 + \frac{3}{5} \left[ \sqrt{5} \tan^{-1}(\sqrt{5}x) \right]_0^1\)
\(= \frac{1}{5} (\log 6 - \log 1) + \frac{3}{\sqrt{5}} [\tan^{-1} \sqrt{5} - \tan^{-1} 0]\)
\(= \frac{1}{5} \log 6 + \frac{3}{\sqrt{5}} \tan^{-1} \sqrt{5}\)

Question. Evaluate \(\int_0^1 x \tan^{-1}(x) dx\)
Answer: Let \(I = \int_0^1 x \tan^{-1} x \, dx\)
Using integration by parts, we get:
\(I = \left[ \tan^{-1} x \cdot \frac{x^2}{2} \right]_0^1 - \int_0^1 \frac{1}{1+x^2} \cdot \frac{x^2}{2} dx\)
\(= \left[ \frac{x^2}{2} \tan^{-1} x \right]_0^1 - \frac{1}{2} \int_0^1 \frac{x^2+1-1}{1+x^2} dx\)
\(= \left[ \frac{x^2}{2} \tan^{-1} x \right]_0^1 - \frac{1}{2} \int_0^1 1 \, dx + \frac{1}{2} \int_0^1 \frac{1}{1+x^2} dx\)
\(= \left[ \frac{x^2}{2} \tan^{-1} x \right]_0^1 - \frac{1}{2} [x]_0^1 + \frac{1}{2} [\tan^{-1} x]_0^1\)
\(= \left[ \frac{1}{2} \tan^{-1}(1) - 0 \right] - \frac{1}{2} (1-0) + \frac{1}{2} [\tan^{-1}(1) - \tan^{-1}(0)]\)
\(= \frac{\pi}{8} - \frac{1}{2} + \frac{1}{2} \left( \frac{\pi}{4} \right)\)
\(= \frac{\pi}{8} - \frac{1}{2} + \frac{\pi}{8} = \frac{\pi}{4} - \frac{1}{2}\)

Question. Evaluate \(\int_{\pi/4}^{\pi/2} e^{2x} \left(\frac{1-\sin 2x}{1-\cos 2x}\right) dx\)
Answer: Let \(I = \int_{\pi/4}^{\pi/2} e^{2x} \left(\frac{1-\sin 2x}{1-\cos 2x}\right) dx\)
First, we simplify the integrand:
\(e^{2x} \left(\frac{1-\sin 2x}{1-\cos 2x}\right) = e^{2x} \left(\frac{1-2\sin x \cos x}{2\sin^2 x}\right) = e^{2x} \left(\frac{1}{2\sin^2 x} - \frac{2\sin x \cos x}{2\sin^2 x}\right)\)
\(= \frac{1}{2} e^{2x} (\csc^2 x - 2\cot x)\)
Now, \(I = \frac{1}{2} \int_{\pi/4}^{\pi/2} e^{2x} \csc^2 x \, dx - \int_{\pi/4}^{\pi/2} e^{2x} \cot x \, dx\)
Integrating the first term by parts (taking \(e^{2x}\) as the first function and \(\csc^2 x\) as the second function):
\(\frac{1}{2} \int e^{2x} \csc^2 x \, dx = \frac{1}{2} \left[ e^{2x} (-\cot x) - \int 2 e^{2x} (-\cot x) dx \right] = -\frac{1}{2} e^{2x} \cot x + \int e^{2x} \cot x \, dx\)
Substituting this back, we get:
\(I = \left[ -\frac{1}{2} e^{2x} \cot x \right]_{\pi/4}^{\pi/2} = -\frac{1}{2} \left[ e^{\pi} \cot\left(\frac{\pi}{2}\right) - e^{\pi/2} \cot\left(\frac{\pi}{4}\right) \right]\)
\(= -\frac{1}{2} \left[ 0 - e^{\pi/2} (1) \right] = \frac{1}{2} e^{\pi/2}\)

Question. Find \(\int_0^1 \frac{\tan^{-1} x}{1+x^2} dx\)
Answer: Let \(I = \int_0^1 \frac{\tan^{-1} x}{1+x^2} dx\)
Let \(\tan^{-1} x = t \Rightarrow \frac{dx}{1+x^2} = dt\).
When \(x = 0\), \(t = \tan^{-1}(0) = 0\).
When \(x = 1\), \(t = \tan^{-1}(1) = \frac{\pi}{4}\).
So, \(I = \int_0^{\pi/4} t \, dt = \left[ \frac{t^2}{2} \right]_0^{\pi/4} = \frac{1}{2} \left[ \left(\frac{\pi}{4}\right)^2 - 0 \right] = \frac{\pi^2}{32}\)

Question. Evaluate \(\int_{-4}^0 |x+2| dx\)
Answer: Let \(I = \int_{-4}^0 |x+2| dx\).
We know that:
\(|x+2| = \begin{cases} -(x+2), & x < -2 \\ x+2, & x \ge -2 \end{cases}\)
Therefore, we split the integral at \(x = -2\):
\(I = \int_{-4}^{-2} -(x+2) dx + \int_{-2}^0 (x+2) dx\)
\(= -\left[ \frac{x^2}{2} + 2x \right]_{-4}^{-2} + \left[ \frac{x^2}{2} + 2x \right]_{-2}^0\)
\(= -\left[ \left(\frac{(-2)^2}{2} + 2(-2)\right) - \left(\frac{(-4)^2}{2} + 2(-4)\right) \right] + \left[ 0 - \left(\frac{(-2)^2}{2} + 2(-2)\right) \right]\)
\(= -\left[ (2 - 4) - (8 - 8) \right] + \left[ 0 - (2 - 4) \right]\)
\(= -[-2] + [2] = 2 + 2 = 4\)

Question. Evaluate \(\int_0^{\pi/4} \frac{\sin 2x}{\sin^4 x + \cos^4 x} dx\)
Answer: Let \(I = \int_0^{\pi/4} \frac{\sin 2x}{\sin^4 x + \cos^4 x} dx = \int_0^{\pi/4} \frac{2\sin x \cos x}{\sin^4 x + \cos^4 x} dx\).
Dividing the numerator and denominator by \(\cos^4 x\), we get:
\(I = \int_0^{\pi/4} \frac{2\tan x \sec^2 x}{\tan^4 x + 1} dx\).
Let \(\tan^2 x = t \Rightarrow 2\tan x \sec^2 x \, dx = dt\).
When \(x = 0\), \(t = \tan^2 0 = 0\).
When \(x = \frac{\pi}{4}\), \(t = \tan^2\left(\frac{\pi}{4}\right) = 1\).
Now, \(I = \int_0^1 \frac{dt}{t^2+1} = \left[ \tan^{-1} t \right]_0^1 = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4}\)

Question. Evaluate \(\int_{-1}^2 |x^3-x| dx\)
Answer: Let \(I = \int_{-1}^2 |x^3-x| dx = \int_{-1}^2 |x(x-1)(x+1)| dx\).
The sign of \((x^3-x)\) in the interval \([-1, 2]\) is as follows:
For \(-1 \le x \le 0\), \(x^3-x \ge 0\).
For \(0 \le x \le 1\), \(x^3-x \le 0\).
For \(1 \le x \le 2\), \(x^3-x \ge 0\).
So, \(I = \int_{-1}^0 (x^3-x) dx + \int_0^1 -(x^3-x) dx + \int_1^2 (x^3-x) dx\)
\(= \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_{-1}^0 - \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_0^1 + \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_1^2\)
\(= -\left( \frac{1}{4} - \frac{1}{2} \right) - \left( \frac{1}{4} - \frac{1}{2} \right) + \left( \left(4 - 2\right) - \left(\frac{1}{4} - \frac{1}{2}\right) \right)\)
\(= \frac{1}{4} + \frac{1}{4} + 2 + \frac{1}{4} = \frac{11}{4}\)

Question. Evaluate \(\int_{-\pi/3}^{\pi/3} \frac{1}{1+e^{\sin x}} dx\)
Answer: Let \(I = \int_{-\pi/3}^{\pi/3} \frac{1}{1+e^{\sin x}} dx\) — (i)
Using the property \(\int_a^b f(x) dx = \int_a^b f(a+b-x) dx\), we get:
\(I = \int_{-\pi/3}^{\pi/3} \frac{1}{1+e^{\sin(-\pi/3+\pi/3-x)}} dx = \int_{-\pi/3}^{\pi/3} \frac{1}{1+e^{\sin(-x)}} dx\)
\(I = \int_{-\pi/3}^{\pi/3} \frac{1}{1+e^{-\sin x}} dx = \int_{-\pi/3}^{\pi/3} \frac{e^{\sin x}}{e^{\sin x}+1} dx\) — (ii)
Adding equations (i) and (ii), we get:
\(2I = \int_{-\pi/3}^{\pi/3} \frac{1+e^{\sin x}}{1+e^{\sin x}} dx = \int_{-\pi/3}^{\pi/3} 1 \, dx\)
\(2I = [x]_{-\pi/3}^{\pi/3} = \frac{\pi}{3} - \left(-\frac{\pi}{3}\right) = \frac{2\pi}{3}\)
\(I = \frac{\pi}{3}\)

Question. Evaluate \(\int_0^{\pi/2} \frac{\sin^3 x}{\sin^3 x + \cos^3 x} dx\)
Answer: Let \(I = \int_0^{\pi/2} \frac{\sin^3 x}{\sin^3 x + \cos^3 x} dx\) — (i)
Using the property \(\int_0^a f(x) dx = \int_0^a f(a-x) dx\), we have:
\(I = \int_0^{\pi/2} \frac{\sin^3(\pi/2 - x)}{\sin^3(\pi/2 - x) + \cos^3(\pi/2 - x)} dx = \int_0^{\pi/2} \frac{\cos^3 x}{\cos^3 x + \sin^3 x} dx\) — (ii)
Adding equations (i) and (ii), we get:
\(2I = \int_0^{\pi/2} \frac{\sin^3 x + \cos^3 x}{\sin^3 x + \cos^3 x} dx = \int_0^{\pi/2} 1 \, dx\)
\(2I = [x]_0^{\pi/2} = \frac{\pi}{2} - 0 = \frac{\pi}{2}\)
\(I = \frac{\pi}{4}\)

Question. Evaluate \(\int_0^{\pi/2} \frac{\tan^7 x}{\cot^7 x + \tan^7 x} dx\)
Answer: Let \(I = \int_0^{\pi/2} \frac{\tan^7 x}{\cot^7 x + \tan^7 x} dx\) — (i)
Using the property \(\int_0^a f(x) dx = \int_0^a f(a-x) dx\), we get:
\(I = \int_0^{\pi/2} \frac{\tan^7(\pi/2 - x)}{\cot^7(\pi/2 - x) + \tan^7(\pi/2 - x)} dx = \int_0^{\pi/2} \frac{\cot^7 x}{\tan^7 x + \cot^7 x} dx\) — (ii)
Adding equations (i) and (ii), we get:
\(2I = \int_0^{\pi/2} \frac{\tan^7 x + \cot^7 x}{\tan^7 x + \cot^7 x} dx = \int_0^{\pi/2} 1 \, dx\)
\(2I = [x]_0^{\pi/2} = \frac{\pi}{2}\)
\(I = \frac{\pi}{4}\)

Long Answer Type Questions

Question. Evaluate \(\int \frac{2x-5}{(2x-3)^3} e^{2x} dx\)
Answer: Let \(I = \int \frac{2x-5}{(2x-3)^3} e^{2x} dx = \int \frac{(2x-3-2)e^{2x}}{(2x-3)^3} dx\)
\(= \int \left[ \frac{1}{(2x-3)^2} - \frac{2}{(2x-3)^3} \right] e^{2x} dx = \int e^{2x}(2x-3)^{-2} dx - 2 \int e^{2x}(2x-3)^{-3} dx\)
Integrating the first term by parts taking \((2x-3)^{-2}\) as the first function:
\(\int e^{2x}(2x-3)^{-2} dx = (2x-3)^{-2} \left( \frac{e^{2x}}{2} \right) - \int -2(2x-3)^{-3} \cdot 2 \left( \frac{e^{2x}}{2} \right) dx\)
\(= \frac{e^{2x}}{2(2x-3)^2} + 2 \int e^{2x}(2x-3)^{-3} dx\)
Substituting this back, we get:
\(I = \left[ \frac{e^{2x}}{2(2x-3)^2} + 2 \int e^{2x}(2x-3)^{-3} dx \right] - 2 \int e^{2x}(2x-3)^{-3} dx\)
\(= \frac{e^{2x}}{2(2x-3)^2} + C\)

Question. Find \(\int e^{2x} \sin(3x+1) dx\)
Answer: Let \(I = \int e^{2x} \sin(3x+1) dx\) — (i)
Integrating by parts taking \(\sin(3x+1)\) as the first function:
\(I = \sin(3x+1) \left(\frac{e^{2x}}{2}\right) - \int 3 \cos(3x+1) \left(\frac{e^{2x}}{2}\right) dx\)
\(I = \frac{e^{2x} \sin(3x+1)}{2} - \frac{3}{2} \int e^{2x} \cos(3x+1) dx\)
Integrating by parts again:
\(\int e^{2x} \cos(3x+1) dx = \cos(3x+1) \left(\frac{e^{2x}}{2}\right) - \int -3 \sin(3x+1) \left(\frac{e^{2x}}{2}\right) dx = \frac{e^{2x}\cos(3x+1)}{2} + \frac{3}{2} I\)
Substituting this back:
\(I = \frac{e^{2x}\sin(3x+1)}{2} - \frac{3}{2} \left[ \frac{e^{2x}\cos(3x+1)}{2} + \frac{3}{2} I \right]\)
\(I = \frac{e^{2x}\sin(3x+1)}{2} - \frac{3e^{2x}\cos(3x+1)}{4} - \frac{9}{4} I\)
\(I \left(1 + \frac{9}{4}\right) = e^{2x} \left[ \frac{\sin(3x+1)}{2} - \frac{3\cos(3x+1)}{4} \right] + C_1\)
\(\frac{13}{4} I = \frac{e^{2x}}{4} [2\sin(3x+1) - 3\cos(3x+1)] + C_1\)
\(I = \frac{e^{2x}}{13} [2\sin(3x+1) - 3\cos(3x+1)] + C\) (where \(C = \frac{4C_1}{13}\))

Question. Evaluate \(\int_0^{\pi/4} \frac{dx}{\cos^3 x \sqrt{2\sin 2x}}\)
Answer: Let \(I = \int_0^{\pi/4} \frac{dx}{\cos^3 x \sqrt{2\sin 2x}}\)
Since \(\sin 2x = 2 \sin x \cos x\), we have:
\(I = \int_0^{\pi/4} \frac{dx}{\cos^3 x \sqrt{4\sin x \cos x}} = \int_0^{\pi/4} \frac{dx}{2 \cos^3 x \cos^{1/2} x \sin^{1/2} x}\)
\(= \frac{1}{2} \int_0^{\pi/4} \frac{dx}{\cos^{7/2} x \sin^{1/2} x} = \frac{1}{2} \int_0^{\pi/4} \frac{dx}{\cos^4 x \left(\frac{\sin x}{\cos x}\right)^{1/2}}\)
\(= \frac{1}{2} \int_0^{\pi/4} \frac{\sec^4 x}{\sqrt{\tan x}} dx = \frac{1}{2} \int_0^{\pi/4} \frac{\sec^2 x (1+\tan^2 x)}{\sqrt{\tan x}} dx\)
Let \(\tan x = t \Rightarrow \sec^2 x \, dx = dt\).
When \(x = 0\), \(t = 0\); when \(x = \frac{\pi}{4}\), \(t = 1\).
\(I = \frac{1}{2} \int_0^1 \left(\frac{1+t^2}{t^{1/2}}\right) dt = \frac{1}{2} \int_0^1 (t^{-1/2} + t^{3/2}) dt\)
\(= \frac{1}{2} \left[ 2t^{1/2} + \frac{2}{5} t^{5/2} \right]_0^1 = \frac{1}{2} \left[ 2 + \frac{2}{5} \right] = 1 + \frac{1}{5} = \frac{6}{5}\)

Question. Evaluate \(\int_0^{\pi/4} \frac{\sin x + \cos x}{16 + 9\sin 2x} dx\)
Answer: Let \(I = \int_0^{\pi/4} \frac{\sin x + \cos x}{16 + 9\sin 2x} dx\).
We write \(\sin 2x = 1 - (\sin x - \cos x)^2\).
Let \(\sin x - \cos x = t \Rightarrow (\cos x + \sin x) dx = dt\).
When \(x = 0\), \(t = \sin 0 - \cos 0 = -1\).
When \(x = \frac{\pi}{4}\), \(t = \sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) = 0\).
Now, \(16 + 9\sin 2x = 16 + 9(1-t^2) = 25-9t^2\).
Thus, \(I = \int_{-1}^0 \frac{dt}{25-9t^2} = \int_{-1}^0 \frac{dt}{5^2 - (3t)^2}\)
Using the standard formula \(\int \frac{du}{a^2-u^2} = \frac{1}{2a} \log\left|\frac{a+u}{a-u}\right|\):
\(I = \left[ \frac{1}{3 \cdot 2 \cdot 5} \log\left|\frac{5+3t}{5-3t}\right| \right]_{-1}^0 = \frac{1}{30} \left[ \log\left|\frac{5+3t}{5-3t}\right| \right]_{-1}^0\)
\(= \frac{1}{30} \left[ \log 1 - \log\left|\frac{5-3}{5+3}\right| \right] = \frac{1}{30} \left[ 0 - \log\left(\frac{2}{8}\right) \right]\)
\(= -\frac{1}{30} \log\left(\frac{1}{4}\right) = \frac{1}{30} \log 4\)

Question. Evaluate \(\int_0^{\pi} x \log|\sin x| dx\)
Answer: Let \(I = \int_0^{\pi} x \log|\sin x| dx\) — (i)
Using the property \(\int_0^a f(x) dx = \int_0^a f(a-x) dx\):
\(I = \int_0^{\pi} (\pi-x) \log|\sin(\pi-x)| dx = \int_0^{\pi} (\pi-x) \log|\sin x| dx\) — (ii)
Adding (i) and (ii):
\(2I = \int_0^{\pi} \pi \log|\sin x| dx \Rightarrow I = \frac{\pi}{2} \int_0^{\pi} \log|\sin x| dx\)
Since \(\sin x \ge 0\) on \([0, \pi]\), \(|\sin x| = \sin x\).
Using the property \(\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx\) if \(f(2a-x) = f(x)\), since \(\log(\sin(\pi-x)) = \log(\sin x)\):
\(I = \frac{\pi}{2} \cdot 2 \int_0^{\pi/2} \log(\sin x) dx = \pi \int_0^{\pi/2} \log(\sin x) dx\)
Using the standard result \(\int_0^{\pi/2} \log(\sin x) dx = -\frac{\pi}{2} \log 2\), we get:
\(I = \pi \left( -\frac{\pi}{2} \log 2 \right) = -\frac{\pi^2}{2} \log 2\)

Question. Evaluate \(\int_0^{\pi/2} \{2 \log(\sin x) - \log(\sin 2x)\} dx\)
Answer: Let \(I = \int_0^{\pi/2} \{2 \log(\sin x) - \log(\sin 2x)\} dx\)
Using logarithmic properties, the integrand can be simplified as:
\(2 \log(\sin x) - \log(\sin 2x) = \log(\sin^2 x) - \log(2 \sin x \cos x) = \log\left(\frac{\sin^2 x}{2 \sin x \cos x}\right) = \log\left(\frac{\tan x}{2}\right) = \log(\tan x) - \log 2\)
So, \(I = \int_0^{\pi/2} \log(\tan x) dx - \int_0^{\pi/2} \log 2 \, dx\)
Let \(I_1 = \int_0^{\pi/2} \log(\tan x) dx\) — (i)
Using \(\int_0^a f(x) dx = \int_0^a f(a-x) dx\):
\(I_1 = \int_0^{\pi/2} \log(\tan(\pi/2 - x)) dx = \int_0^{\pi/2} \log(\cot x) dx\) — (ii)
Adding (i) and (ii):
\(2I_1 = \int_0^{\pi/2} [\log(\tan x) + \log(\cot x)] dx = \int_0^{\pi/2} \log(\tan x \cdot \cot x) dx = \int_0^{\pi/2} \log 1 \, dx = 0\)
Thus, \(I_1 = 0\).
Therefore,
\(I = 0 - \log 2 \int_0^{\pi/2} 1 \, dx = -\log 2 [x]_0^{\pi/2} = -\frac{\pi}{2} \log 2\)