CBSE Class 12 Mathematics Integrals Class Test Set 01

Class Test Questions with Solutions

Question 1. If \( \frac{d}{dx}f(x) = 4x^3 - \frac{3}{x^4} \) such that f(2) = 0, then find f(x). [1]
Answer:
Given \( \frac{d}{dx}f(x) = 4x^3 - \frac{3}{x^4} \)
\( f(x) = \int \left(4x^3 - \frac{3}{x^4}\right) dx \) [integrating on both sides]
\( f(x) = x^4 + \frac{1}{x^3} + C \) ...(1)
Put x = 2:
\( f(2) = 16 + \frac{1}{8} + C \) [Given f(2) = 0]
\( 0 = \frac{129 + 8C}{8} \Rightarrow C = -\frac{129}{8} \)
Putting \( C = -\frac{129}{8} \) in equation (1),
\( f(x) = x^4 + \frac{1}{x^3} - \frac{129}{8} \)

Question 2. \( \int_{-1}^{2} |x|\,dx \) is equal to: [1]
Answer:
\( I = \int_{-1}^{2} |x|\,dx \)
∵ \( \int_{a}^{b} f(x)\,dx = \int_{a}^{c} f(x)\,dx + \int_{c}^{b} f(x)\,dx \)
\( I = -\int_{-1}^{0} x\,dx + \int_{0}^{2} x\,dx = -\left[\frac{x^2}{2}\right]_{-1}^{0} + \left[\frac{x^2}{2}\right]_{0}^{2} \)
\( I = -\left[0 - \frac{1}{2}\right] + \left[\frac{4}{2} - 0\right] = \frac{1}{2} + \frac{4}{2} \)
\( I = \frac{5}{2} \)

Question 3. Evaluate: \( \int \frac{1}{\sin x\cos^3 x}\,dx \) [1]
Answer:
Let \( I = \int \frac{dx}{\sin x\cos^3 x} \)
[Dividing numerator and denominator by \( \cos^4 x \)]
\( I = \int \frac{\sec^4 x}{\tan x}\,dx = \int \frac{(1 + \tan^2 x)\sec^2 x}{\tan x}\,dx \)
Put tan x = t ⇒ \( \sec^2 x\,dx = dt \)
\( I = \int \frac{(1 + t^2)}{t}\,dt = \int \left(\frac{1}{t} + t\right) dt \)
\( I = \log|t| + \frac{t^2}{2} + C = \log|\tan x| + \frac{\tan^2 x}{2} + C \)

Question 4. Evaluate: \( \int \sqrt{25 - x^2}\,dx \) [1]
Answer:
\( I = \int \sqrt{25 - x^2}\,dx = \frac{x}{2}\sqrt{25 - x^2} + \frac{25}{2}\sin^{-1}\frac{x}{5} + C \)
\( \left(\because \int \sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C\right) \)

Question 5. If \( \int_{0}^{a} \frac{1}{4 + x^2}\,dx = \frac{\pi}{8} \), find the value of a. [2]
Answer:
\( \int_{0}^{a} \frac{1}{4 + x^2}\,dx = \frac{\pi}{8} \)
⇒ \( \left[\frac{1}{2}\tan^{-1}\frac{x}{2}\right]_{0}^{a} = \frac{\pi}{8} \) \( \left(\because \int \frac{dx}{a^2 + x^2} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C\right) \)
⇒ \( \frac{1}{2}\tan^{-1}\frac{a}{2} = \frac{\pi}{8} \Rightarrow \frac{a}{2} = \tan\frac{\pi}{4} = 1 \Rightarrow a = 2 \)

Question 6. Evaluate: \( \int_{0}^{\pi} \frac{4x\sin x}{1 + \cos^2 x}\,dx \) [3]
Answer:
\( I = \int_{0}^{\pi} \frac{4x\sin x}{1 + \cos^2 x}\,dx \)
\( I = \int_{0}^{\pi} \frac{4(\pi - x)\sin(\pi - x)}{1 + \cos^2(\pi - x)}\,dx \) \( \left[\text{Applying } \int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a - x)\,dx\right] \)
\( I = \int_{0}^{\pi} \frac{4\pi\sin x}{1 + \cos^2 x}\,dx - \int_{0}^{\pi} \frac{4x\sin x}{1 + \cos^2 x}\,dx = \int_{0}^{\pi} \frac{4\pi\sin x}{1 + \cos^2 x}\,dx - I \)
\( 2I = 4\pi\int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x}\,dx \)
\( 2I = 4\pi \cdot 2 \times \int_{0}^{\pi/2} \frac{\sin x}{1 + \cos^2 x}\,dx \) \( \left\{\text{Applying } \int_{0}^{2a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx \text{ if } f(2a - x) = f(x)\right\} \)
\( I = 4\pi\int_{0}^{\pi/2} \frac{\sin x}{1 + \cos^2 x}\,dx \)
Put cos x = t ⇒ −sin x dx = dt; when x = 0, t = 1 and when \( x = \frac{\pi}{2} \), t = 0
∴ \( I = 4\pi\int_{1}^{0} \frac{-dt}{1 + t^2} = 4\pi\int_{0}^{1} \frac{dt}{1 + t^2} \) \( \left\{\int_{a}^{b} f(x)\,dx = -\int_{b}^{a} f(x)\,dx\right\} \)
\( I = 4\pi\left[\tan^{-1} t\right]_{0}^{1} = 4\pi\left[\tan^{-1} 1 - \tan^{-1} 0\right] \)
\( I = 4\pi \times \frac{\pi}{4} = \pi^2 \)

Question 7. Find: \( \int \frac{x^2 + x + 1}{(x + 2)(x^2 + 1)}\,dx \) [3]
Answer:
The integrand is a proper rational function. Decompose the rational function into partial fractions. Write
\( \frac{x^2 + x + 1}{(x^2 + 1)(x + 2)} = \frac{A}{x + 2} + \frac{Bx + C}{(x^2 + 1)} \)
Therefore, \( x^2 + x + 1 = A(x^2 + 1) + (Bx + C)(x + 2) \)
Equating the coefficients of \( x^2 \), x and of the constant term on both sides, we get A + B = 1, 2B + C = 1 and A + 2C = 1. Solving these equations, we get \( A = \frac{3}{5},\ B = \frac{2}{5} \) and \( C = \frac{1}{5} \).
Thus, the integrand is given by
\( \frac{x^2 + x + 1}{(x^2 + 1)(x + 2)} = \frac{3}{5(x + 2)} + \frac{\frac{2}{5}x + \frac{1}{5}}{x^2 + 1} = \frac{3}{5(x + 2)} + \frac{1}{5}\left(\frac{2x + 1}{x^2 + 1}\right) \)
Therefore, \( \int \frac{x^2 + x + 1}{(x^2 + 1)(x + 2)}\,dx = \frac{3}{5}\int \frac{dx}{x + 2} + \frac{1}{5}\int \frac{2x}{x^2 + 1}\,dx + \frac{1}{5}\int \frac{1}{x^2 + 1}\,dx \)
\( I = \frac{3}{5}\log|x + 2| + \frac{1}{5}\log|x^2 + 1| + \frac{1}{5}\tan^{-1} x + C \)

Question 8. Evaluate: \( \int \frac{1 - x^2}{x(1 - 2x)}\,dx \) [3]
Answer:
Let \( I = \int \frac{1 - x^2}{x(1 - 2x)}\,dx \)
Here the integrand is an improper rational function,
so \( \frac{1 - x^2}{x(1 - 2x)} = \frac{1}{2} + \frac{\left(1 - \frac{x}{2}\right)}{x(1 - 2x)} \) [Dividing numerator by denominator]
Now \( I = \int \left(\frac{1}{2} + \frac{\left(1 - \frac{x}{2}\right)}{x(1 - 2x)}\right) dx = \frac{1}{2}\int 1\,dx + \int \frac{\left(1 - \frac{x}{2}\right)}{x(1 - 2x)}\,dx \)
Let \( I = I_1 + I_2 \), where \( I_1 = \frac{1}{2}\int 1\,dx = \frac{1}{2}x + C_1 \) and \( I_2 = \int \frac{\left(1 - \frac{x}{2}\right)}{x(1 - 2x)}\,dx \)
Let \( \frac{1 - \frac{x}{2}}{x(1 - 2x)} = \frac{A}{x} + \frac{B}{1 - 2x} \)
⇒ \( -\frac{x}{2} + 1 = A(1 - 2x) + Bx \)
On comparing coefficients of x on both sides, we get \( -2A + B = -\frac{1}{2} \) ...(1)
On comparing constant terms on both sides, we get A = 1
Putting A = 1 in eq. (1), we get \( B = -\frac{1}{2} + 2 \times 1 = \frac{3}{2} \)
Now \( \frac{1 - \frac{x}{2}}{x(1 - 2x)} = \frac{1}{x} + \frac{3}{2} \cdot \frac{1}{(1 - 2x)} \)
⇒ \( I_2 = \int \frac{1}{x}\,dx + \frac{3}{2}\int \frac{dx}{1 - 2x} = \log|x| - \frac{3}{4}\log|1 - 2x| + C_2 \)
\( I = \frac{1}{2}x + \log|x| - \frac{3}{4}\log|1 - 2x| + C \), where \( C = C_1 + C_2 \)

Question 9. Evaluate: \( \int_{0}^{\pi/2} \frac{x}{\sin x + \cos x}\,dx \) [3]
Answer:
\( I = \int_{0}^{\pi/2} \frac{x}{\sin x + \cos x}\,dx \) ...(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) \( \left[\because \int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a - x)\,dx\right] \)
Adding equations (1) and (2), we get
⇒ \( 2I = \int_{0}^{\pi/2} \frac{\frac{\pi}{2}}{\cos x + \sin x}\,dx = \frac{\pi}{2}\int_{0}^{\pi/2} \frac{1}{\sin x + \cos x}\,dx \)
\( I = \frac{\pi}{4\sqrt{2}}\int_{0}^{\pi/2} \frac{1}{\sin\left(x + \frac{\pi}{4}\right)}\,dx = \frac{\pi}{4\sqrt{2}}\int_{0}^{\pi/2} \text{cosec}\left(x + \frac{\pi}{4}\right) dx \)
\( I = \frac{\pi}{4\sqrt{2}}\left[\log\left(\text{cosec}\left(x + \frac{\pi}{4}\right) - \cot\left(x + \frac{\pi}{4}\right)\right)\right]_{0}^{\pi/2} \)
\( I = \frac{\pi}{4\sqrt{2}}\left[\log\left(\text{cosec}\left(\frac{\pi}{2} + \frac{\pi}{4}\right) - \cot\left(\frac{\pi}{2} + \frac{\pi}{4}\right)\right) - \log\left(\text{cosec}\frac{\pi}{4} - \cot\frac{\pi}{4}\right)\right] \)
\( I = \frac{\pi}{4\sqrt{2}}\left[\log\left(\sec\frac{\pi}{4} + \tan\frac{\pi}{4}\right) - \log\left(\text{cosec}\frac{\pi}{4} - \cot\frac{\pi}{4}\right)\right] \)
\( I = \frac{\pi}{4\sqrt{2}}\left[\log(\sqrt{2} + 1) - \log(\sqrt{2} - 1)\right] \)
\( I = \frac{\pi}{4\sqrt{2}}\log\frac{\sqrt{2} + 1}{\sqrt{2} - 1} \)

Question 10. Find: \( \int \left[\sqrt{\cot x} + \sqrt{\tan x}\right] dx \) [5]
Answer:
Let \( I = \int (\sqrt{\cot x} + \sqrt{\tan x})\,dx = \int \frac{\sin x + \cos x}{\sqrt{\sin x\cos x}}\,dx \)
\( = \sqrt{2}\int \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}}\,dx \)
Put sin x − cos x = t, (cos x + sin x) dx = dt
\( = \sqrt{2}\int \frac{dt}{\sqrt{1 - t^2}} = \sqrt{2}\left[\sin^{-1} t\right] \)
\( = \sqrt{2}\left[\sin^{-1}(\sin x - \cos x)\right] + C \)
Aliter:
We have \( I = \int \left[\sqrt{\cot x} + \sqrt{\tan x}\right] dx = \int \sqrt{\tan x}(1 + \cot x)\,dx \)
Put \( \tan x = t^2 \), so that \( \sec^2 x\,dx = 2t\,dt \) or \( dx = \frac{2t\,dt}{1 + t^4} \)
Then \( I = \int t\left(1 + \frac{1}{t^2}\right)\frac{2t}{(1 + t^4)}\,dt = 2\int \frac{(t^2 + 1)}{t^4 + 1}\,dt = 2\int \frac{\left(1 + \frac{1}{t^2}\right)}{\left(t^2 + \frac{1}{t^2}\right)}\,dt = 2\int \frac{\left(1 + \frac{1}{t^2}\right) dt}{\left(t - \frac{1}{t}\right)^2 + 2} \)
Put \( t - \frac{1}{t} = y \), so that \( \left(1 + \frac{1}{t^2}\right) dt = dy \). Then
\( I = 2\int \frac{dy}{y^2 + (\sqrt{2})^2} = \sqrt{2}\tan^{-1}\frac{y}{\sqrt{2}} + C = \sqrt{2}\tan^{-1}\frac{\left(t - \frac{1}{t}\right)}{\sqrt{2}} + C \)
\( = \sqrt{2}\tan^{-1}\left(\frac{t^2 - 1}{\sqrt{2}t}\right) + C = \sqrt{2}\tan^{-1}\left(\frac{\tan x - 1}{\sqrt{2\tan x}}\right) + C \)

Question 11. Evaluate the definite integral: \( \int_{0}^{\pi/4} \frac{\sin x + \cos x}{16 + 9\sin 2x}\,dx \) [5]
Answer:
Let \( I = \int_{0}^{\pi/4} \frac{\sin x + \cos x}{16 + 9\sin 2x}\,dx = \int_{0}^{\pi/4} \frac{\sin x + \cos x}{16 + 9\{1 - (\sin x - \cos x)^2\}}\,dx \)
Also, let sin x − cos x = t ⇒ (cos x + sin x) dx = dt
When x = 0, t = −1 and when \( x = \frac{\pi}{4} \), t = 0
⇒ \( (\sin x - \cos x)^2 = t^2 \Rightarrow \sin^2 x + \cos^2 x - 2\sin x\cos x = t^2 \Rightarrow 1 - \sin 2x = t^2 \Rightarrow \sin 2x = 1 - t^2 \)
∴ \( I = \int_{-1}^{0} \frac{dt}{16 + 9(1 - t^2)} = \int_{-1}^{0} \frac{dt}{16 + 9 - 9t^2} = \int_{-1}^{0} \frac{dt}{25 - 9t^2} \)
\( = \frac{1}{9}\int_{-1}^{0} \frac{dt}{\frac{25}{9} - t^2} = \frac{1}{9}\int_{-1}^{0} \frac{dt}{\left(\frac{5}{3}\right)^2 - (t)^2} \)
\( = \frac{1}{9} \cdot \frac{1}{2} \times \frac{3}{5}\left[\log\left(\frac{\frac{5}{3} + t}{\frac{5}{3} - t}\right)\right]_{-1}^{0} \)
\( = \frac{1}{30}\left[\log(1) - \log\left(\frac{1}{4}\right)\right] = \frac{1}{30}\log 4 \)