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Vidaara.orgClass 12 · Mathematics
CodeVID-M12-07-CT
Integrals — Full Chapter Test
Chapter: Integrals
Topic: Complete chapter — all topics
Maximum Marks: 35
Time: 80 minutes
Name: ____________________ Roll No.: __________ Date: ____________

General Instructions

  • This is a full-length, CBSE board-hard test covering the whole chapter. All questions are compulsory.
  • Section A carries 1 mark each (including an assertion-reason item), Section B 2 marks, Section C 3 marks and Section D 5 marks (long answer / case based).
  • Show all working for Sections B, C and D. Final answers are given in the Answer Key.
Section A — Multiple Choice Questions 5 × 1 = 5 marks
1.
$\displaystyle\int e^x\left(\tan x+\sec^2x\right)dx$ equals:
  • A.$e^x\tan x+C$
  • B.$e^x\sec x+C$
  • C.$e^x\sec^2x+C$
  • D.$e^x(\tan x-1)+C$
2.
The value of $\displaystyle\int_{-\pi/2}^{\pi/2}\left(x^3+x\cos x+\tan^5x+1\right)dx$ is:
  • A.$0$
  • B.$\pi$
  • C.$2\pi$
  • D.$\dfrac{\pi}{2}$
3.
Assertion (A): $\displaystyle\int_0^{\pi/2}\dfrac{\sin x}{\sin x+\cos x}\,dx=\dfrac{\pi}{4}$.
Reason (R): $\displaystyle\int_0^af(x)\,dx=\int_0^af(a-x)\,dx$.
  • A.Both A and R are true and R is the correct explanation of A
  • B.Both A and R are true but R is not the correct explanation of A
  • C.A is true but R is false
  • D.A is false but R is true
4.
$\displaystyle\int\dfrac{dx}{\sqrt{9-4x^2}}$ equals:
  • A.$\dfrac13\sin^{-1}\dfrac{2x}{3}+C$
  • B.$\dfrac12\sin^{-1}\dfrac{2x}{3}+C$
  • C.$\sin^{-1}\dfrac{2x}{3}+C$
  • D.$\dfrac12\sin^{-1}\dfrac{x}{3}+C$
5.
The value of $\displaystyle\int_1^4|x-2|\,dx$ is:
  • A.$3$
  • B.$\dfrac52$
  • C.$\dfrac72$
  • D.$2$
Section B — Short Answer I 4 × 2 = 8 marks
6.
Evaluate $\displaystyle\int\dfrac{\cos2x-\cos2\alpha}{\cos x-\cos\alpha}\,dx$.
7.
Evaluate $\displaystyle\int_{-\pi/4}^{\pi/4}\sin^2x\,dx$.
8.
Evaluate $\displaystyle\int\dfrac{dx}{e^x+e^{-x}}$.
9.
Evaluate $\displaystyle\int_0^1xe^{x^2}\,dx$.
Section C — Short Answer II 4 × 3 = 12 marks
10.
Evaluate $\displaystyle\int\dfrac{x^2+1}{(x-1)^2(x+3)}\,dx$.
11.
Evaluate $\displaystyle\int x^2\tan^{-1}x\,dx$.
12.
Evaluate $\displaystyle\int_0^{\pi}\dfrac{x}{1+\sin x}\,dx$.
13.
Evaluate $\displaystyle\int_0^{\pi/2}\left(2\log\sin x-\log\sin2x\right)dx$.
Section D — Long Answer / Case Based 2 × 5 = 10 marks
14.
Evaluate $\displaystyle\int_0^{\pi/2}\dfrac{\sin x\cos x}{\sin^4x+\cos^4x}\,dx$.
15.
(a) Evaluate $I=\displaystyle\int_0^{\pi/2}\dfrac{dx}{1+\sqrt{\tan x}}$. [3]
(b) Hence, or otherwise, show that $\displaystyle\int_0^{\pi/2}\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx=\dfrac{\pi}{4}$. [2]

Answer Key

Section A — Multiple Choice Questions
  1. (A) $e^x\tan x+C$
  2. (B) $\pi$
  3. (A) Both A and R are true and R is the correct explanation of A
  4. (B) $\dfrac12\sin^{-1}\dfrac{2x}{3}+C$
  5. (B) $\dfrac52$
Section B — Short Answer I
  1. $\cos2x-\cos2\alpha=2(\cos^2x-\cos^2\alpha)=2(\cos x-\cos\alpha)(\cos x+\cos\alpha)$. So the integral is $2\int(\cos x+\cos\alpha)dx=2\sin x+2x\cos\alpha+C$.
  2. $\displaystyle\int_{-\pi/4}^{\pi/4}\dfrac{1-\cos2x}{2}dx=\left[\dfrac x2-\dfrac{\sin2x}{4}\right]_{-\pi/4}^{\pi/4}=\left(\dfrac{\pi}{8}-\dfrac14\right)-\left(-\dfrac{\pi}{8}+\dfrac14\right)=\dfrac{\pi}{4}-\dfrac12$.
  3. $\displaystyle\int\dfrac{e^x}{e^{2x}+1}dx$. Put $t=e^x$: $\displaystyle\int\dfrac{dt}{1+t^2}=\tan^{-1}t+C=\tan^{-1}(e^x)+C$.
  4. Put $t=x^2$, $x\,dx=\dfrac{dt}{2}$: $\displaystyle\dfrac12\int_0^1e^t\,dt=\dfrac{e-1}{2}$.
Section C — Short Answer II
  1. Write $\dfrac{x^2+1}{(x-1)^2(x+3)}=\dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{x+3}$. Then $x=1$ gives $B=\dfrac12$; $x=-3$ gives $C=\dfrac58$; comparing $x^2$ coefficients, $A+C=1\Rightarrow A=\dfrac38$. Hence the integral is $\dfrac38\log|x-1|-\dfrac{1}{2(x-1)}+\dfrac58\log|x+3|+C$.
  2. By parts: $\dfrac{x^3}{3}\tan^{-1}x-\dfrac13\displaystyle\int\dfrac{x^3}{1+x^2}dx$. Since $\dfrac{x^3}{1+x^2}=x-\dfrac{x}{1+x^2}$, $\displaystyle\int\dfrac{x^3}{1+x^2}dx=\dfrac{x^2}{2}-\dfrac12\log(1+x^2)$. So the result is $\dfrac{x^3}{3}\tan^{-1}x-\dfrac{x^2}{6}+\dfrac16\log(1+x^2)+C$.
  3. Let $I=\displaystyle\int_0^\pi\dfrac{x}{1+\sin x}dx=\int_0^\pi\dfrac{\pi-x}{1+\sin x}dx$. Adding, $2I=\pi\displaystyle\int_0^\pi\dfrac{dx}{1+\sin x}$. Write $1+\sin x=1+\cos\left(x-\frac{\pi}{2}\right)=2\cos^2\left(\frac x2-\frac{\pi}{4}\right)$, so $\displaystyle\int_0^\pi\dfrac{dx}{1+\sin x}=\left[\tan\left(\frac x2-\frac{\pi}{4}\right)\right]_0^\pi=\tan\frac{\pi}{4}-\tan\left(-\frac{\pi}{4}\right)=2$. So $2I=2\pi$ and $I=\pi$.
  4. $2\log\sin x-\log(2\sin x\cos x)=\log\sin x-\log\cos x-\log2$. Since $\displaystyle\int_0^{\pi/2}\log\sin x\,dx=\int_0^{\pi/2}\log\cos x\,dx$, those terms cancel and the value is $-\dfrac{\pi}{2}\log2$.
Section D — Long Answer / Case Based
  1. Put $t=\sin^2x$, $dt=2\sin x\cos x\,dx$; limits $0\to1$. Since $\sin^4x+\cos^4x=t^2+(1-t)^2=2t^2-2t+1$, the integral is $\dfrac12\displaystyle\int_0^1\dfrac{dt}{2t^2-2t+1}=\dfrac14\int_0^1\dfrac{dt}{\left(t-\frac12\right)^2+\frac14}=\dfrac14\cdot2\left[\tan^{-1}(2t-1)\right]_0^1=\dfrac12\left(\dfrac{\pi}{4}+\dfrac{\pi}{4}\right)=\dfrac{\pi}{4}$.
  2. (a) $I=\displaystyle\int_0^{\pi/2}\dfrac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx$. Using $\displaystyle\int_0^af(x)dx=\int_0^af(a-x)dx$: $I=\displaystyle\int_0^{\pi/2}\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$. Adding the two forms, $2I=\displaystyle\int_0^{\pi/2}1\,dx=\dfrac{\pi}{2}$, so $I=\dfrac{\pi}{4}$. (b) The second form of $I$ above is exactly this integral, so its value is $\dfrac{\pi}{4}$.
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