CBSE Class 12 Mathematics Integrals Previous Year Questions Set 01

Very Short Answer Questions (1 Mark)

Question 1. If \( \int_{0}^{a} \frac{1}{4 + x^2}\,dx = \frac{\pi}{8} \), find the value of a. [CBSE 2014, 1M]
Answer:
\( \int_{0}^{a} \frac{1}{4 + x^2}\,dx = \frac{\pi}{8} \)
⇒ \( \left[\frac{1}{2}\tan^{-1}\frac{x}{2}\right]_{0}^{a} = \frac{\pi}{8} \)
or \( \frac{1}{2}\tan^{-1}\frac{a}{2} = \frac{\pi}{8} \)
⇒ \( \frac{a}{2} = \tan\frac{\pi}{4} \)
a = 2 [1]

Question 2. Find: \( \int \frac{\sin^2 x - \cos^2 x}{\sin x \cos x}\,dx \) [CBSE 2017, 1M]
Answer:
\( \int \frac{\sin^2 x - \cos^2 x}{\sin x \cos x}\,dx = \int \frac{\sin^2 x}{\sin x \cos x}\,dx - \int \frac{\cos^2 x}{\sin x \cos x}\,dx \)
\( = \int \tan x\,dx - \int \cot x\,dx \)
\( = \log|\sec x| - \log|\sin x| + C \) [1]

Question 3. Evaluate: \( \int_{-\pi/2}^{\pi/2} x\cos^2 x\,dx \) [CBSE 2020, 1M]
Answer:
Let \( I = \int_{-\pi/2}^{\pi/2} x \cdot \cos^2 x\,dx \)
Here \( f(x) = x \cdot \cos^2 x \)
\( f(-x) = (-x)\cos^2(-x) = -x\cos^2 x \)
f(−x) = −f(x), i.e. f(x) is an odd function
∴ I = 0 \( \left(\because \int_{-a}^{a} f(x)\,dx = 0 \text{ if f(x) is an odd function}\right) \) [1]

Short Answer Questions (2 Marks)

Question 4. Find: \( \int \frac{dx}{5 - 8x - x^2} \) [CBSE 2017, 2M]
Answer:
\( \int \frac{dx}{5 - 8x - x^2} = \int \frac{dx}{-[x^2 + 8x - 5 + 16 - 16]} \)
\( = \int \frac{dx}{-[(x + 4)^2 - 21]} \)
\( = \int \frac{dx}{(\sqrt{21})^2 - (x + 4)^2} \) [1]
\( = \frac{1}{2\sqrt{21}}\log\left|\frac{\sqrt{21} + (x + 4)}{\sqrt{21} - (x + 4)}\right| + C \) [1]

Question 5. Evaluate: \( \int \frac{\cos 2x + 2\sin^2 x}{\cos^2 x}\,dx \) [CBSE 2018, 2M]
Answer:
\( \int \frac{\cos 2x + 2\sin^2 x}{\cos^2 x}\,dx = \int \frac{1 - 2\sin^2 x + 2\sin^2 x}{\cos^2 x}\,dx \) [1]
\( = \int \frac{1}{\cos^2 x}\,dx = \int \sec^2 x\,dx = \tan x + C \) [1]

Question 6. Find: \( \int \frac{x - 5}{(x - 3)^3}e^x\,dx \) [CBSE 2019, 2M]
Answer:
\( \int \frac{x - 5}{(x - 3)^3}e^x\,dx = \int e^x\left[\frac{(x - 3) - 2}{(x - 3)^3}\right] dx \) [½]
\( = \int e^x\left[\frac{(x - 3)}{(x - 3)^3} - \frac{2}{(x - 3)^3}\right] dx \)
\( = \int e^x\left[\frac{1}{(x - 3)^2} + \left\{\frac{-2}{(x - 3)^3}\right\}\right] dx \) [½]
\( = \frac{e^x}{(x - 3)^2} + C \) [1]
\( \left[\because \int e^x[f(x) + f'(x)]\,dx = e^x f(x) + C\right] \)

Question 7. Find: \( \int \frac{\sin^3 x + \cos^3 x}{\sin^2 x\cos^2 x}\,dx \)
OR
Find: \( \int \frac{x - 3}{(x - 1)^3}e^x\,dx \) [CBSE 2019, 2M]
Answer:
\( \int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cdot \cos^2 x}\,dx = \int \left(\frac{\sin x}{\cos^2 x} + \frac{\cos x}{\sin^2 x}\right) dx \)
\( = \int (\tan x\sec x + \text{cosec}\,x \cdot \cot x)\,dx \) [1]
\( = \sec x - \text{cosec}\,x + C \) [1]
OR
\( \int \frac{x - 3}{(x - 1)^3}e^x\,dx = \int e^x\left(\frac{x - 1 - 2}{(x - 1)^3}\right) dx \)
\( = \int e^x\left(\frac{1}{(x - 1)^2} + \left(\frac{-2}{(x - 1)^3}\right)\right) dx \) [1]
\( = \frac{e^x}{(x - 1)^2} + C \) \( \left[\int e^x(f(x) + f'(x))\,dx = e^x f(x) + C\right] \) [1]

Question 8. Find: \( \int \frac{\tan^3 x}{\cos^3 x}\,dx \) [CBSE 2020, 2M]
Answer:
Let \( I = \int \frac{\tan^3 x}{\cos^3 x}\,dx = \int \frac{\sin^3 x}{\cos^6 x}\,dx = \int \frac{\sin^2 x \cdot \sin x}{\cos^6 x}\,dx \)
\( = \int \frac{(1 - \cos^2 x) \cdot \sin x}{\cos^6 x}\,dx \) [1]
Put cos x = t ⇒ −sin x dx = dt ⇒ sin x dx = −dt
Now \( I = -\int \frac{(1 - t^2)}{t^6}\,dt = -\int \left(\frac{1}{t^6} - \frac{1}{t^4}\right) dt \)
\( = -\left[\frac{-1}{5t^5} + \frac{1}{3t^3}\right] + C \)
\( = \frac{1}{5\cos^5 x} - \frac{1}{3\cos^3 x} + C \) [1]

Long Answer Questions I (4 Marks)

Question 9. Evaluate: \( \int_{0}^{\pi/2} \log\sin x\,dx \) [CBSE 2008, 4M]
Answer:
Let \( I = \int_{0}^{\pi/2} \log\sin x\,dx \) ...(i)
\( I = \int_{0}^{\pi/2} \log\sin\left(\frac{\pi}{2} - x\right) dx \) \( \left[\because \int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a - x)\,dx\right] \)
\( I = \int_{0}^{\pi/2} \log\cos x\,dx \) ...(ii) [1]
On adding (i) and (ii), we get
\( 2I = \int_{0}^{\pi/2} (\log\sin x + \log\cos x)\,dx = \int_{0}^{\pi/2} \log(\sin x\cos x)\,dx \)
\( 2I = \int_{0}^{\pi/2} \log\frac{\sin 2x}{2}\,dx \) [∵ sin 2x = 2 sin x cos x] [½]
\( 2I = \int_{0}^{\pi/2} (\log\sin 2x - \log 2)\,dx = \int_{0}^{\pi/2} \log\sin 2x\,dx - \int_{0}^{\pi/2} \log 2\,dx \) [½]
Put 2x = t ⇒ 2 dx = dt
When x = 0, t = 0; when \( x = \frac{\pi}{2} \), \( t = 2 \cdot \frac{\pi}{2} = \pi \) [½]
∴ \( 2I = \int_{0}^{\pi} \log\sin t\,\frac{dt}{2} - \log 2[x]_{0}^{\pi/2} \)
\( \left[\int_{0}^{2a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx \text{ if } f(2a - x) = f(x)\right] \) and ∵ log sin(π − t) = log sin t
∴ \( 2I = \frac{1}{2} \cdot 2\int_{0}^{\pi/2} \log\sin t\,dt - \log 2\left[\frac{\pi}{2} - 0\right] \) [½]
\( 2I = \int_{0}^{\pi/2} \log\sin t\,dt - \frac{\pi}{2}\log 2 \)
\( 2I = \int_{0}^{\pi/2} \log\sin x\,dx - \frac{\pi}{2}\log 2 \) \( \left[\int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(t)\,dt\right] \)
\( 2I = I - \frac{\pi}{2}\log 2 \)
∴ \( I = -\frac{\pi}{2}\log 2 \) [1]

Question 10. Evaluate: \( \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{\tan x}} \) [CBSE 2011, 4M]
Answer:
\( I = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{\tan x}} = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx \) ...(1) [1]
\( I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}}\,dx \) ...(2) \( \left[\text{Using } \int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a + b - x)\,dx\right] \) [1]
Adding equations (1) and (2),
\( 2I = \int_{\pi/6}^{\pi/3} dx \Rightarrow 2I = \frac{\pi}{6} \Rightarrow I = \frac{\pi}{12} \) [2]

Question 11. Evaluate: \( \int_{-1}^{2} |x^3 - x|\,dx \) [CBSE 2012, 4M]
Answer:
We note that \( x^3 - x \geq 0 \) on [−1, 0]; \( x^3 - x \leq 0 \) on [0, 1] and \( x^3 - x \geq 0 \) on [1, 2].
So by the property of definite integrals, we get
\( \int_{-1}^{2} |x^3 - x|\,dx = \int_{-1}^{0} (x^3 - x)\,dx + \int_{0}^{1} -(x^3 - x)\,dx + \int_{1}^{2} (x^3 - x)\,dx \) [1]
\( = \int_{-1}^{0} (x^3 - x)\,dx + \int_{0}^{1} (x - x^3)\,dx + \int_{1}^{2} (x^3 - x)\,dx = \left[\frac{x^4}{4} - \frac{x^2}{2}\right]_{-1}^{0} + \left[\frac{x^2}{2} - \frac{x^4}{4}\right]_{0}^{1} + \left[\frac{x^4}{4} - \frac{x^2}{2}\right]_{1}^{2} \) [1]
\( = -\left(\frac{1}{4} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + (4 - 2) - \left(\frac{1}{4} - \frac{1}{2}\right) = -\frac{1}{4} + \frac{1}{2} + \frac{1}{2} - \frac{1}{4} + 2 - \frac{1}{4} + \frac{1}{2} = \frac{3}{2} - \frac{3}{4} + 2 = \frac{11}{4} \) [2]

Question 12. Evaluate: \( \int_{0}^{2\pi} \frac{1}{1 + e^{\sin x}}\,dx \) [CBSE 2013, 4M]
Answer:
\( I = \int_{0}^{2\pi} \frac{1}{1 + e^{\sin x}}\,dx \) ...(i)
\( I = \int_{0}^{2\pi} \frac{1}{1 + e^{-\sin x}}\,dx \) \( \left(\text{By } \int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a - x)\,dx\right) \) [1]
\( = \int_{0}^{2\pi} \frac{e^{\sin x}}{1 + e^{\sin x}}\,dx \) ...(ii) [1]
From (i) + (ii),
\( 2I = \int_{0}^{2\pi} \left[\frac{1 + e^{\sin x}}{1 + e^{\sin x}}\right] dx \) [1]
\( 2I = \int_{0}^{2\pi} dx \Rightarrow 2I = (x)_{0}^{2\pi} \) [½]
\( 2I = 2\pi \Rightarrow I = \pi \) [½]

Question 13. Evaluate: \( \int \frac{\sin x - x\cos x}{x(x + \sin x)}\,dx \)
OR
Evaluate: \( \int \frac{x^3}{(x - 1)(x^2 + 1)}\,dx \) [CBSE 2015, 4M]
Answer:
Let \( I = \int \frac{\sin x - x\cos x}{x(x + \sin x)}\,dx = \int \left[\frac{\sin x}{x(x + \sin x)} - \frac{x\cos x}{x(x + \sin x)}\right] dx \)
\( I = \int \left[\frac{(x + \sin x) - x}{x(x + \sin x)} - \frac{\cos x}{x + \sin x}\right] dx \) [1]
\( I = \int \left[\frac{1}{x} - \frac{1}{x + \sin x} - \frac{\cos x}{x + \sin x}\right] dx \) [1]
\( I = \int \frac{1}{x}\,dx - \int \frac{1 + \cos x}{x + \sin x}\,dx \) [1]
\( I = \log|x| - \log|x + \sin x| + C \) \( \left[\because \int \frac{f'(x)}{f(x)}\,dx = \log f(x) + C\right] \)
\( I = \log\left|\frac{x}{x + \sin x}\right| + C \) [1]
OR
Let \( I = \int \frac{x^3}{(x - 1)(x^2 + 1)}\,dx \)
\( \frac{x^3}{(x - 1)(x^2 + 1)} = 1 + \frac{x^2 - x + 1}{(x - 1)(x^2 + 1)} \)
Let \( \frac{x^2 - x + 1}{(x - 1)(x^2 + 1)} = \frac{A}{(x - 1)} + \frac{Bx + C}{x^2 + 1} \) [1]
\( x^2 - x + 1 = A(x^2 + 1) + (x - 1)(Bx + C) \) ...(i)
Using (i), we have:
When x = 1, \( A = \frac{1}{2} \)
and x = 0, \( 1 = A - C \Rightarrow C = A - 1 = -\frac{1}{2} \)
On equating coefficients of \( x^2 \): \( 1 = A + B \Rightarrow B = 1 - A = \frac{1}{2} \) [1]
∴ \( \frac{x^2 - x + 1}{(x - 1)(x^2 + 1)} = \frac{1}{2}\left(\frac{1}{x - 1}\right) + \frac{1}{2}\left(\frac{x - 1}{x^2 + 1}\right) \)
\( I = \int \left[1 + \frac{1}{2}\left(\frac{1}{x - 1}\right) + \frac{1}{2}\left(\frac{x - 1}{x^2 + 1}\right)\right] dx \)
\( = \int \left[1 + \frac{1}{2}\left(\frac{1}{x - 1}\right) + \frac{1}{2} \times \frac{1}{2}\left(\frac{2x}{x^2 + 1}\right) - \frac{1}{2}\left(\frac{1}{x^2 + 1}\right)\right] dx \) [1]
\( = x + \frac{1}{2}\log|x - 1| + \frac{1}{4}\log(x^2 + 1) - \frac{1}{2}\tan^{-1} x + C \)
\( = x + \frac{1}{4}\log\left[(x - 1)^2(x^2 + 1)\right] - \frac{1}{2}\tan^{-1} x + C \) [1]

Question 14. Evaluate: \( \int_{0}^{\pi/2} \frac{\cos^2 x\,dx}{1 + 3\sin^2 x} \) [CBSE 2015, 4M]
Answer:
\( I = \int_{0}^{\pi/2} \frac{\cos^2 x}{1 + 3\sin^2 x}\,dx = \int_{0}^{\pi/2} \frac{1}{\sec^2 x + 3\tan^2 x}\,dx \) (dividing by \( \cos^2 x \))
\( = \int_{0}^{\pi/2} \frac{1}{(1 + 4\tan^2 x)} \times \frac{\sec^2 x}{(1 + \tan^2 x)}\,dx \) [1]
Put \( \tan x = t \Rightarrow \sec^2 x\,dx = dt \); when x = 0, t = 0 and x = π/2, t → ∞
\( I = \int_{0}^{\infty} \frac{1}{(1 + t^2)(1 + 4t^2)}\,dt = -\frac{1}{3}\int_{0}^{\infty} \frac{1}{1 + t^2}\,dt + \frac{4}{3}\int_{0}^{\infty} \frac{1}{1 + (2t)^2}\,dt \) [1]
\( = -\frac{1}{3}\left[\tan^{-1} t\right]_{0}^{\infty} + \frac{4}{3} \times \frac{1}{2}\left[\tan^{-1} 2t\right]_{0}^{\infty} \) [1]
\( = -\frac{1}{3}\left(\frac{\pi}{2}\right) + \frac{2}{3} \times \left(\frac{\pi}{2}\right) = -\frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{6} \) [1]

Question 15. Evaluate: \( \int_{0}^{\pi/4} \left(\frac{\sin x + \cos x}{3 + \sin 2x}\right) dx \) [CBSE 2015, 4M]
Answer:
Let \( I = \int_{0}^{\pi/4} \frac{\sin x + \cos x}{3 + \sin 2x}\,dx = \int_{0}^{\pi/4} \frac{\sin x + \cos x}{4 - (\sin x - \cos x)^2}\,dx \) [1]
Put sin x − cos x = t ⇒ (cos x + sin x) dx = dt
when x = 0, t = −1 and \( x = \frac{\pi}{4} \), t = 0
⇒ \( I = \int_{-1}^{0} \frac{dt}{(2)^2 - t^2} \) [1]
or \( I = \frac{1}{4}\left[\log\frac{2 + t}{2 - t}\right]_{-1}^{0} \) [1]
\( I = \frac{1}{4}\left[\log 1 - \log\frac{1}{3}\right] \)
\( I = -\frac{1}{4}\log\frac{1}{3} = \frac{1}{4}\log 3 \) [1]

Question 16. Evaluate: \( \int_{-2}^{2} \frac{x^2}{1 + 5^x}\,dx \) [CBSE 2016, 4M]
Answer:
Let \( I = \int_{-2}^{2} \frac{x^2}{1 + 5^x}\,dx \) ...(i)
⇒ \( I = \int_{-2}^{2} \frac{x^2}{1 + 5^{-x}}\,dx \) \( \left[\because \int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a + b - x)\,dx\right] \) [1]
or \( I = \int_{-2}^{2} \frac{x^2\,5^x}{1 + 5^x}\,dx \) ...(ii) [1]
Equation (i) + (ii) gives
\( 2I = \int_{-2}^{2} x^2\,dx \)
⇒ \( 2I = 2\int_{0}^{2} x^2\,dx \) \( \left[\because \int_{-a}^{a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx, \text{ if f(x) is an even function}\right] \) [1]
⇒ \( I = \int_{0}^{2} x^2\,dx = \frac{1}{3}(x^3)_{0}^{2} = \frac{1}{3}(2^3 - 0^3) \)
∴ \( I = \frac{8}{3} \) [1]

Question 17. Evaluate: \( \int \frac{(2x - 5)e^{2x}}{(2x - 3)^3}\,dx \)
OR
Evaluate: \( \int \frac{x^2 + x + 1}{(x^2 + 1)(x + 2)}\,dx \) [CBSE 2016, 4M]
Answer:
\( I = \int e^{2x}\frac{(2x - 5)}{(2x - 3)^3}\,dx = \int e^{2x}\left[\frac{(2x - 3)}{(2x - 3)^3} - \frac{2}{(2x - 3)^3}\right] dx \)
\( I = \int e^{2x}\left[\frac{1}{(2x - 3)^2} - \frac{2}{(2x - 3)^3}\right] dx \) [1]
Put 2x = t, \( dx = \frac{dt}{2} \)
\( I = \frac{1}{2}\int e^t\left[\frac{1}{(t - 3)^2} - \frac{2}{(t - 3)^3}\right] dt \)
\( I = \frac{1}{2} \cdot \frac{e^t}{(t - 3)^2} + C \) \( \left[\because \int e^x[f(x) + f'(x)]\,dx = e^x f(x) + C\right] \) [2]
\( I = \frac{1}{2} \cdot \frac{e^{2x}}{(2x - 3)^2} + C \) [1]
OR
\( I = \int \frac{x^2 + x + 1}{(x^2 + 1)(x + 2)}\,dx \)
Let \( \frac{x^2 + x + 1}{(x + 2)(x^2 + 1)} = \frac{A}{x + 2} + \frac{Bx + C}{x^2 + 1} \) ...(1)
\( x^2 + x + 1 = A(x^2 + 1) + (Bx + C)(x + 2) \)
At x = −2, \( A = \frac{3}{5} \)
and at x = 0, \( A + 2C = 1 \Rightarrow C = \frac{1}{5} \)
Equating the coefficients of \( x^2 \): \( 1 = A + B \Rightarrow B = 1 - \frac{3}{5} = \frac{2}{5} \) [1]
Putting these in equation (1) and integrating,
∴ \( I = \int \frac{x^2 + x + 1}{(x + 2)(x^2 + 1)}\,dx = \frac{3}{5}\int \frac{1}{x + 2}\,dx + \int \frac{\frac{2}{5}x + \frac{1}{5}}{x^2 + 1}\,dx \)
\( = \frac{3}{5}\int \frac{1}{x + 2}\,dx + \frac{1}{5}\int \frac{2x}{x^2 + 1}\,dx + \frac{1}{5}\int \frac{1}{x^2 + 1}\,dx \) [2]
\( = \frac{3}{5}\log|x + 2| + \frac{1}{5}\log(x^2 + 1) + \frac{1}{5}\tan^{-1} x + C \) [1]

Question 18. Evaluate: \( \int_{1}^{4} \{|x - 1| + |x - 2| + |x - 4|\}\,dx \) [CBSE 2017, 4M]
Answer:
Let \( I = \int_{1}^{4} \{|x - 1| + |x - 2| + |x - 4|\}\,dx \)
\( I = \int_{1}^{2} [(x - 1) - (x - 2) - (x - 4)]\,dx + \int_{2}^{4} [(x - 1) + (x - 2) - (x - 4)]\,dx \) [2]
\( I = \int_{1}^{2} (5 - x)\,dx + \int_{2}^{4} (x + 1)\,dx \)
\( I = \left(5x - \frac{x^2}{2}\right)_{1}^{2} + \left(\frac{x^2}{2} + x\right)_{2}^{4} \) [1]
\( I = \left[(10 - 2) - \left(5 - \frac{1}{2}\right)\right] + [(8 + 4) - (2 + 2)] \)
\( I = \frac{7}{2} + 8 = \frac{23}{2} \) [1]

Question 19. Evaluate: \( \int \frac{\sin\theta\,d\theta}{(4 + \cos^2\theta)(2 - \sin^2\theta)} \) [CBSE 2017, 4M]
Answer:
\( I = \int \frac{\sin\theta\,d\theta}{(4 + \cos^2\theta)(2 - \sin^2\theta)} = \int \frac{\sin\theta\,d\theta}{(4 + \cos^2\theta)(1 + 1 - \sin^2\theta)} \) [½]
⇒ \( I = \int \frac{\sin\theta\,d\theta}{(4 + \cos^2\theta)(1 + \cos^2\theta)} \)
Let cos θ = t ⇒ −sin θ dθ = dt
⇒ \( I = \int \frac{-dt}{(4 + t^2)(1 + t^2)} \) [1]
\( I = -\frac{1}{3}\int \frac{3}{(4 + t^2)(1 + t^2)}\,dt = -\frac{1}{3}\int \frac{(4 + t^2) - (1 + t^2)}{(4 + t^2)(1 + t^2)}\,dt \) [1]
\( = \frac{1}{3}\left[\int \frac{1}{4 + t^2}\,dt - \int \frac{1}{1 + t^2}\,dt\right] \)
\( = \frac{1}{3}\left[\frac{1}{2}\tan^{-1}\left(\frac{t}{2}\right) - \tan^{-1} t\right] + C \) [1]
\( = \frac{1}{6}\tan^{-1}\left(\frac{\cos\theta}{2}\right) - \frac{1}{3}\tan^{-1}(\cos\theta) + C \) [½]

Question 20. Evaluate: \( \int \frac{2\cos x}{(1 - \sin x)(2 - \cos^2 x)}\,dx \) [CBSE 2018, 2019, 4M]
Answer:
\( I = \int \frac{2\cos x}{(1 - \sin x)(2 - \cos^2 x)}\,dx = \int \frac{2\cos x}{(1 - \sin x)(1 + 1 - \cos^2 x)}\,dx = \int \frac{2\cos x}{(1 - \sin x)(1 + \sin^2 x)}\,dx \)
Put sin x = t ⇒ cos x dx = dt
\( I = \int \frac{2\,dt}{(1 - t)(1 + t^2)} \)
⇒ \( I = -2I_1 \) ...(1) [1]
Now, \( I_1 = \int \frac{dt}{(t - 1)(t^2 + 1)} \)
Let \( \frac{1}{(t - 1)(t^2 + 1)} = \frac{A}{(t - 1)} + \frac{Bt + C}{(t^2 + 1)} \)
⇒ \( 1 = A(t^2 + 1) + Bt(t - 1) + C(t - 1) \)
Putting t = 1: \( 1 = A \times 2 \Rightarrow A = \frac{1}{2} \)
Comparing coefficients of \( t^2 \) on both sides, we get: \( 0 = A + B \Rightarrow B = -\frac{1}{2} \)
Putting t = 0: 1 = A × 1 + C × (−1) ⇒ \( 1 = A - C \Rightarrow C = A - 1 = -\frac{1}{2} \) [1]
⇒ \( I_1 = \frac{1}{2}\int \frac{dt}{t - 1} - \frac{1}{2}\int \frac{t + 1}{t^2 + 1}\,dt \)
\( = \frac{1}{2}\log|t - 1| - \frac{1}{4}\int \frac{2t}{t^2 + 1}\,dt - \frac{1}{2}\int \frac{dt}{t^2 + 1} + C_1 \)
⇒ \( I_1 = \frac{1}{2}\log|t - 1| - \frac{1}{4}\log|t^2 + 1| - \frac{1}{2}\tan^{-1} t + C_1 \) [1]
From equation (1), we have: \( I = -2I_1 = -2\left[\frac{1}{2}\log|t - 1| - \frac{1}{4}\log|t^2 + 1| - \frac{1}{2}\tan^{-1} t + C_1\right] \)
\( = -\log|t - 1| + \frac{1}{2}\log|t^2 + 1| + \tan^{-1} t - 2C_1 \)
⇒ \( I = -\log|\sin x - 1| + \frac{1}{2}\log|1 + \sin^2 x| + \tan^{-1}(\sin x) + C \), where \( C = -2C_1 \) [1]

Question 21. Prove that \( \int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a - x)\,dx \) and hence evaluate \( \int_{0}^{\pi/2} \frac{x}{\sin x + \cos x}\,dx \). [CBSE 2019, 4M]
Answer:
Let \( I = \int_{0}^{a} f(a - x)\,dx \)
Put a − x = t ⇒ dx = −dt
When x = 0, t = a and x = a, t = 0
⇒ \( I = -\int_{a}^{0} f(t)\,dt \)
or \( I = \int_{0}^{a} f(t)\,dt \) \( \left[\because \int_{a}^{b} f(x)\,dx = -\int_{b}^{a} f(x)\,dx\right] \)
⇒ \( I = \int_{0}^{a} f(x)\,dx \) \( \left[\because \int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(t)\,dt\right] \) [1]
Now, \( I = \int_{0}^{\pi/2} \frac{x}{\sin x + \cos x}\,dx \) ...(1)
\( I = \int_{0}^{\pi/2} \frac{\left(\frac{\pi}{2} - x\right)}{\sin\left(\frac{\pi}{2} - x\right) + \cos\left(\frac{\pi}{2} - x\right)}\,dx \) ...(2)
(1) + (2) gives: \( 2I = \int_{0}^{\pi/2} \frac{\frac{\pi}{2}}{\cos x + \sin x}\,dx \)
∴ \( 2I = \frac{\pi}{2}\int_{0}^{\pi/2} \frac{1}{\sin x + \cos x}\,dx \)
⇒ \( I = \frac{\pi}{4\sqrt{2}}\int_{0}^{\pi/2} \frac{1}{\sin\left(x + \frac{\pi}{4}\right)}\,dx \)
or \( I = \frac{\pi}{4\sqrt{2}}\int_{0}^{\pi/2} \text{cosec}\left(x + \frac{\pi}{4}\right) dx \) [1]
⇒ \( I = \frac{\pi}{4\sqrt{2}}\left[\log\left(\text{cosec}\left(x + \frac{\pi}{4}\right) - \cot\left(x + \frac{\pi}{4}\right)\right)\right]_{0}^{\pi/2} \) [1]
or \( I = \frac{\pi}{4\sqrt{2}}\left[\log\left(\text{cosec}\left(\frac{\pi}{2} + \frac{\pi}{4}\right) - \cot\left(\frac{\pi}{2} + \frac{\pi}{4}\right)\right) - \log\left(\text{cosec}\frac{\pi}{4} - \cot\frac{\pi}{4}\right)\right] \)
∴ \( I = \frac{\pi}{4\sqrt{2}}\left[\log\left(\sec\frac{\pi}{4} + \tan\frac{\pi}{4}\right) - \log\left(\text{cosec}\frac{\pi}{4} - \cot\frac{\pi}{4}\right)\right] \)
or \( I = \frac{\pi}{4\sqrt{2}}\left[\log(\sqrt{2} + 1) - \log(\sqrt{2} - 1)\right] \)
⇒ \( I = \frac{\pi}{4\sqrt{2}}\log\frac{\sqrt{2} + 1}{\sqrt{2} - 1} \) [1]

Question 22. Find: \( \int \frac{x^3 + 1}{x^3 - x}\,dx \) [CBSE 2020, 4M]
Answer:
\( I = \int \frac{x^3 + 1 - x + x}{x^3 - x}\,dx \) or \( I = \int \left(1 + \frac{1 + x}{x^3 - x}\right) dx \) [1]
\( = \int \left(1 + \frac{1 + x}{x(x^2 - 1)}\right) dx = \int \left(1 + \frac{1}{x(x - 1)}\right) dx \) [1]
\( = \int \left(1 + \frac{1}{x - 1} - \frac{1}{x}\right) dx \)
\( = x + \log|x - 1| - \log|x| + C \) [1]
\( = x + \log\left|\frac{x - 1}{x}\right| + C \) [1]

Long Answer Questions II (6 Marks)

Question 23. Evaluate: \( \int_{0}^{\pi} \frac{x\,dx}{a^2\cos^2 x + b^2\sin^2 x} \) [CBSE 2009, 6M]
Answer:
\( I = \int_{0}^{\pi} \frac{x\,dx}{a^2\cos^2 x + b^2\sin^2 x} \)
\( I = \int_{0}^{\pi} \frac{(\pi - x)\,dx}{a^2\cos^2 x + b^2\sin^2 x} \) \( \left[\because \int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a - x)\,dx\right] \) [1]
Adding, \( 2I = \int_{0}^{\pi} \frac{\pi\,dx}{a^2\cos^2 x + b^2\sin^2 x} \) [1]
\( I = \frac{\pi}{2}\int_{0}^{\pi} \frac{dx}{a^2\cos^2 x + b^2\sin^2 x} \)
\( I = \frac{\pi}{2} \times 2\int_{0}^{\pi/2} \frac{dx}{a^2\cos^2 x + b^2\sin^2 x} \) \( \left[\because \int_{0}^{2a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx, \text{ if } f(2a - x) = f(x)\right] \) [1]
\( I = \pi\int_{0}^{\pi/2} \frac{\sec^2 x\,dx}{a^2 + b^2\tan^2 x} \) [1]
Put tan x = t ⇒ \( \sec^2 x\,dx = dt \); when x = 0 ⇒ t = 0 and x = π/2 ⇒ t → ∞
\( I = \pi\int_{0}^{\infty} \frac{dt}{a^2 + b^2t^2} = \frac{\pi}{b^2}\int_{0}^{\infty} \frac{dt}{t^2 + \left(\frac{a}{b}\right)^2} \) [1]
\( I = \left(\frac{\pi}{b^2}\right)\frac{1}{\left(\frac{a}{b}\right)}\left[\tan^{-1}\left(\frac{t}{a/b}\right)\right]_{0}^{\infty} = \frac{\pi}{ab} \cdot \frac{\pi}{2} = \frac{\pi^2}{2ab} \) [1]

Question 24. Evaluate: \( \int \frac{1}{\cos^4 x + \sin^4 x}\,dx \) [CBSE 2014, 6M]
Answer:
\( \int \frac{dx}{\cos^4 x + \sin^4 x} = \int \frac{\frac{1}{\cos^4 x}}{1 + \tan^4 x}\,dx = \int \frac{\sec^2 x \cdot \sec^2 x}{1 + \tan^4 x}\,dx = \int \frac{(1 + \tan^2 x)\sec^2 x}{1 + \tan^4 x}\,dx \)
Put tan x = t ⇒ \( \sec^2 x\,dx = dt \) [2]
\( = \int \frac{(1 + t^2)\,dt}{1 + t^4} = \int \frac{\left(\frac{1}{t^2} + 1\right) dt}{\frac{1}{t^2} + t^2} = \int \frac{\left(1 + \frac{1}{t^2}\right) dt}{\left(t - \frac{1}{t}\right)^2 + 2} \)
Put \( t - \frac{1}{t} = z \Rightarrow \left(1 + \frac{1}{t^2}\right) dt = dz \) [2]
\( = \int \frac{dz}{z^2 + (\sqrt{2})^2} = \frac{1}{\sqrt{2}}\tan^{-1}\frac{z}{\sqrt{2}} + C \)
\( = \frac{1}{\sqrt{2}}\tan^{-1}\frac{1}{\sqrt{2}}\left(t - \frac{1}{t}\right) + C = \frac{1}{\sqrt{2}}\tan^{-1}\frac{1}{\sqrt{2}}\left(\tan x - \frac{1}{\tan x}\right) + C \)
\( = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\tan x - \cot x}{\sqrt{2}}\right) + C \) [2]

Question 25. Evaluate: \( \int (\sqrt{\cot x} + \sqrt{\tan x})\,dx \) [CBSE 2014, 6M]
Answer:
Let \( I = \int (\sqrt{\cot x} + \sqrt{\tan x})\,dx = \int \frac{\sin x + \cos x}{\sqrt{\sin x\cos x}}\,dx \)
\( = \sqrt{2}\int \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}}\,dx \) [2]
Put sin x − cos x = t ⇒ (cos x + sin x) dx = dt
\( = \sqrt{2}\int \frac{dt}{\sqrt{1 - t^2}} \) [2]
\( = \sqrt{2}\left[\sin^{-1} t\right] = \sqrt{2}\left[\sin^{-1}(\sin x - \cos x)\right] + C \) [2]
Aliter:
We have \( I = \int \left[\sqrt{\cot x} + \sqrt{\tan x}\right] dx = \int \sqrt{\tan x}(1 + \cot x)\,dx \)
Put \( \tan x = t^2 \Rightarrow \sec^2 x\,dx = 2t\,dt \Rightarrow dx = \frac{2t\,dt}{1 + t^4} \) [1]
Then \( I = \int t\left(1 + \frac{1}{t^2}\right)\frac{2t}{(1 + t^4)}\,dt = 2\int \frac{(t^2 + 1)}{t^4 + 1}\,dt = 2\int \frac{\left(1 + \frac{1}{t^2}\right) dt}{\left(t^2 + \frac{1}{t^2}\right)} = 2\int \frac{\left(1 + \frac{1}{t^2}\right) dt}{\left(t - \frac{1}{t}\right)^2 + 2} \) [2]
Put \( t - \frac{1}{t} = y \Rightarrow \left(1 + \frac{1}{t^2}\right) dt = dy \) [1]
\( I = 2\int \frac{dy}{y^2 + (\sqrt{2})^2} = \sqrt{2}\tan^{-1}\frac{y}{\sqrt{2}} + C = \sqrt{2}\tan^{-1}\frac{\left(t - \frac{1}{t}\right)}{\sqrt{2}} + C \)
\( = \sqrt{2}\tan^{-1}\left(\frac{t^2 - 1}{\sqrt{2}t}\right) + C = \sqrt{2}\tan^{-1}\left(\frac{\tan x - 1}{\sqrt{2\tan x}}\right) + C \) [2]

Question 26. Evaluate: \( \int \frac{1}{\sin^4 x + \sin^2 x\cos^2 x + \cos^4 x}\,dx \) [CBSE 2014, 6M]
Answer:
\( I = \int \frac{1}{\sin^4 x + \sin^2 x\cos^2 x + \cos^4 x}\,dx \)
\( I = \int \frac{\sec^2 x(1 + \tan^2 x)}{\tan^4 x + \tan^2 x + 1}\,dx \) (dividing numerator and denominator by \( \cos^4 x \)) [1]
Put tan x = t ⇒ \( \sec^2 x\,dx = dt \)
\( I = \int \frac{(1 + t^2)}{t^4 + t^2 + 1}\,dt = \int \frac{(1 + 1/t^2)}{t^2 + \frac{1}{t^2} + 1}\,dt = \int \frac{(1 + 1/t^2)}{(t - 1/t)^2 + (\sqrt{3})^2}\,dt \) [2½]
Put \( t - \frac{1}{t} = u \Rightarrow \left(1 + \frac{1}{t^2}\right) dt = du \)
\( I = \int \frac{du}{u^2 + (\sqrt{3})^2} = \frac{1}{\sqrt{3}}\tan^{-1}\frac{u}{\sqrt{3}} + C \)
\( I = \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{\tan^2 x - 1}{\sqrt{3}\tan x}\right) + C \) [2½]

Question 27. Evaluate: \( \int_{0}^{\pi/4} \frac{\sin x + \cos x}{16 + 9\sin 2x}\,dx \) [CBSE 2018, 6M]
Answer:
Let \( I = \int_{0}^{\pi/4} \frac{\sin x + \cos x}{16 + 9\sin 2x}\,dx = \int_{0}^{\pi/4} \frac{\sin x + \cos x}{16 + 9\{1 - (\sin x - \cos x)^2\}}\,dx \)
Put sin x − cos x = t ⇒ (cos x + sin x) dx = dt
when x = 0, t = −1 and \( x = \frac{\pi}{4} \), t = 0 [1]
∴ \( I = \int_{-1}^{0} \frac{dt}{16 + 9(1 - t^2)} \) [1½]
\( = \int_{-1}^{0} \frac{dt}{16 + 9 - 9t^2} = \int_{-1}^{0} \frac{dt}{25 - 9t^2} = \frac{1}{9}\int_{-1}^{0} \frac{dt}{\frac{25}{9} - t^2} \)
\( = \frac{1}{9}\int_{-1}^{0} \frac{dt}{\left(\frac{5}{3}\right)^2 - (t)^2} \) [1]
\( = \frac{1}{9} \cdot \frac{1}{2} \times \frac{3}{5}\left[\log\left(\frac{\frac{5}{3} + t}{\frac{5}{3} - t}\right)\right]_{-1}^{0} \) [1½]
\( = \frac{1}{30}\left[\log(1) - \log\left(\frac{1}{4}\right)\right] \)
\( = \frac{1}{30}\log 4 \) [1]