Integral Calculus for JEE Main & Advanced: Practice Problems

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25 Multiple Choice Questions (MCQs) on Integral Calculus for students of class XI and XII preparing for board examinations or JEE Mains, IIT Advanced WBJEE or any other competitive entrance examination.

👨‍🏫 Author: Vinod Singh

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Advanced Integral Calculus - Definite and Indefinite Integration

Test your understanding of core concepts.Master Integral Calculus for JEE Main & Advanced. Practice hand-picked problems with step-by-step solutions, advanced shortcuts, and integration techniques.

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Question 1

Let \(f(x)=\frac{x}{(1+x^n)^{1/n}}\) for \(n \geq 2\) and \(g(x)=f \circ f \circ \dots \circ f\) (\(n\) times), then \(\int x^{n-2} g(x) dx \) is equal to

A\( \frac{1}{n(n+1)} (1+nx^n)^{1-\frac{1}{n}}+c\) B\( \frac{1}{n(n-1)} (1+nx^n)^{1-\frac{1}{n}}+c\) C\( \frac{1}{n(n-1)} (1+nx^n)^{1+\frac{1}{n}}+c \) D\( \frac{1}{n(n-1)} (1-nx^n)^{1-\frac{1}{n}}+c\)

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Question 2

\(\int_{\frac{1}{2026}}^{2026} \frac{tan^{-1} x}{x} dx \quad =\)

A \(\frac{\pi}{4} \ln 2026\) B \(\sqrt{2}\frac{\pi}{8} \ln 2026\) C \(\frac{\pi}{2} \ln 2026\) D \(\frac{\pi}{3} \ln 2026\)

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Question 3

\(\int \frac{dx}{\sqrt[3]{x}+\sqrt[5]{x}} \quad =\)

A \( = 15 \left( \frac{x^{2/3}}{10} - \frac{x^{8/15}}{8} + \frac{x^{2/5}}{6} - \frac{x^{4/15}}{4} + \frac{x^{2/15}}{2} - \frac{1}{2} \ln |x^{2/15} + 1| \right) + c \) B \( = 15 \left( \frac{x^{2/3}}{10} + \frac{x^{8/15}}{8} + \frac{x^{2/5}}{6} + \frac{x^{4/15}}{4} + \frac{x^{2/15}}{2} - \frac{1}{2} \ln |x^{2/15} + 1| \right) + c \) C \( = 15 \left( -\frac{x^{2/3}}{10} - \frac{x^{8/15}}{8} - \frac{x^{2/5}}{6} - \frac{x^{4/15}}{4} - \frac{x^{2/15}}{2} - \frac{1}{2} \ln |x^{2/15} + 1| \right) + c \) D \( = 17 \left( \frac{x^{2/3}}{10} - \frac{x^{8/15}}{8} + \frac{x^{2/5}}{6} - \frac{x^{4/15}}{4} + \frac{x^{2/15}}{2} - \frac{1}{2} \ln |x^{2/15} + 1| \right) + c \)

Put \(x=z^{15}\)

Question 4

\(\int (x^6+x^3) \sqrt[3]{x^3+2} \quad dx \quad =\)

A\(\frac{1}{2}(x^6+2x^3)^{\frac{3}{4}} +c\) B\(\frac{1}{8}(x^6+2x^3)^{\frac{5}{3}} +c\) C\(\frac{1}{8}(x^6+2x^3)^{\frac{4}{3}} +c\) D\(\frac{1}{8}(x^6+x^3)^{\frac{4}{3}} +c\)

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Question 5

\( \int_{\pi/2}^{5\pi/2} \frac{e^{tan^{-1}(\sin x)}}{e^{tan^{-1}(\sin x)}+e^{tan^{-1}(\cos x)}} dx \quad = \)

A\( 2\pi \) B\( \frac{\pi}{8}\) C\( \frac{\pi}{2} \) D\( \pi \)

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Question 6

\(\int \frac{dx}{\tan x + \sec x + \cot x + \csc x} dx \quad = \)

A\(\frac{1}{2}(\cos x - \sin x -x)+c\) B\(\frac{1}{2}(-\cos x + \sin x -x)+c\) C\(\frac{1}{2}(-\cos x + \sin x + x)+c\) D\(\frac{1}{2}(-\cos x - \sin x -x)+c\)

See that \(\frac{1}{\tan x + \sec x + \cot x + \csc x} = \frac{\sin x \cos x}{1+\sin x +\cos x}=\frac{1}{2}\frac{(\sin x +\cos x)^2-1^2}{1+\sin x +\cos x}\)

Question 7

If \( f \) is an even function and \(I= \int_{0}^{\pi/2}f(\cos 2x) \cos x \quad dx \), then

A \( I= \sqrt{2}\int_{0}^{\pi/4}f(\cos 2x) \cos x \quad dx \) B \( I= \sqrt{2}\int_{0}^{\pi/4}f(\cos 2x) \sin x \quad dx \) C \( I= \sqrt{2}\int_{0}^{\pi/4}f(\sin 2x) \cos x \quad dx \) D \( I= \sqrt{2}\int_{0}^{\pi/4}f(\sin 2x) \sin x \quad dx \)

Use King's Rule

Question 8

Let \( f \) be a polynomial function such that \( f(x^2+1) = x^4+5x^2+2\), for all \( x \in \mathbb{R}.\) Then \(\int_{0}^{3} f(x) dx \) is equal to

A \( \frac{41}{3} \) B \( \frac{33}{2} \) C \( \frac{27}{2} \) D \( \frac{5}{3} \)

See that \(f(x^2+1)=(x^2+1)^2+3(x^2+1)-2\)

Question 9

Evaluate \(\int_{-\pi/3}^{\pi/3} \frac{\pi+4x^3}{2-cos\big(|x|+\frac{\pi}{3}\big)} dx\)

A\(\frac{4\pi}{\sqrt{3}}tan^{-1}(\frac{1}{3})\) B\(\frac{4\pi}{\sqrt{2}}tan^{-1}(\frac{1}{2})\) C\(\frac{4\pi}{\sqrt{3}}tan^{-1}(\frac{1}{2})\) D\(\frac{4\pi}{\sqrt{2}}tan^{-1}(\frac{1}{3})\)

Express \(\frac{\pi+4x^3}{2-cos\big(|x|+\frac{\pi}{3}\big)}\) as a sum of an odd and an even function, then split

Question 10

Lrt \(f: (0,\infty) \rightarrow \mathbb{R}\) and \(F(x)=\int_{0}^{x} f(t) dt.\) If \(F(x^2)=x^2(1+x),\) then \(f(4)\) equals

A\(\frac{5}{4}\) B\(7\) C\(4\) D\(2\)

See that \(F'(x)=f(x)\) by fundamental theorem of integral calculus and \(f(x^2)=F'(x^2)= \frac{3x+2}{2}\)

Question 11

Evaluate \( \int sin^{-1} \big(\frac{2x+2}{\sqrt{4x^2+8x+13}} \big) dx.\)

A\((x+1)tan^{-1}\big(\frac{2x+2}{3}\big)+\frac{3}{4}ln|4x^2+8x+13|+c\) B\((x+1)tan^{-1}\big(\frac{2x+2}{3}\big)-\frac{1}{4}ln|4x^2+8x+13|+c\) C\((x+1)tan^{-1}\big(\frac{2x+2}{3}\big)-\frac{3}{2}ln|4x^2+8x+13|+c\) D\((x+1)tan^{-1}\big(\frac{2x+2}{3}\big)-\frac{3}{4}ln|4x^2+8x+13|+c\)

\(4x^2+8x+13=(2x+2)^2+3^2\) now put \( 2x+2 = t\) and then subsutitute \( t=3 \tan \theta \)

Question 12

Let \(T > 0\) be a fixed real number. Suppose \(f\) is a continuous function such that for all \(x \in \mathbb{R}, f(x+T)=f(x).\) If \(I=\int_{0}^{T} f(x) \\dx\) then the value of \(\int_{3}^{3+3T} f(2x) \\dx\) is

A\(3I\) B\(2I \) C\(\frac{3}{2}I\) D\(6I\)

\(\int_{3}^{3+3T} f(2x) \\dx\) = \(\frac{1}{2}\int_{6}^{6+6T} f(t) \\dt\) on substituting \(2x=t\) now split the integral form 6 to T, T to 2T, ...., 5T to 6T and 6T to 6+6T and think!

Question 13

The integral \(\int_{-1/a}^{1/a} \big( [x]+\ln (\frac{1+x}{1-x})\big) dx\) where \(a>1\) equals

A\(0\) B\(\frac{-1}{a}\) C\(1\) D\(a \ln \frac{1}{a}\)

Find out the odd function!

Question 14

For any natural number \(m\), evaluate \(\int (x^{3m}+x^{2m}+x^m)(2x^{2m}+3x^m+6)^{1/m} dx, x>0\)

A \( \frac{1}{6(m+1)} (2x^{3m}+3x^{2m}+6x^m)^{\frac{m-1}{m}}+c\) B \( \frac{1}{6(m-1)} (2x^{3m}+3x^{2m}+6x^m)^{\frac{m+1}{m}}+c\) C \( \frac{1}{(m+1)} (2x^{3m}+3x^{2m}+6x^m)^{\frac{m+1}{m}}+c\) D \( \frac{1}{6(m+1)} (2x^{3m}+3x^{2m}+6x^m)^{\frac{m+1}{m}}+c\)

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Question 15

Let \(f(x)= \frac{e^x}{1+e^x} \quad I_1 = \int_{f(-a)}^{f(a)}xg(x(1-x)) dx \) and \(I_2 = \int_{f(-a)}^{f(a)} g(x(1-x)) dx, \) then the value of \(\frac{I_2}{I_1}\) is

A\(3\) B\(4\) C\(2\) D\(-1\)

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Question 16

Evaluate \(\int_{0}^{2\pi} |1+2\sin x| dx \)

A\(\frac{2\pi}{3} (4+\sqrt{3})\) B\(\frac{4\pi}{3} (2+\sqrt{3})\) C\(\frac{\pi}{3} (4+\sqrt{3})\) D\(\frac{\pi}{3} (2+\sqrt{3})\)

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Question 17

The integral \(\int \frac{2x^{12}+5x^9}{(x^5+x^3+1)^3} dx \)

A\(\frac{-x^5}{(x^5+x^3+1)^{2}}+c\) B\(\frac{x^{10}}{2(x^5+x^3+1)^{2}}+c\) C\(\frac{x^5}{2(x^5+x^3+1)^{2}}+c\) D\(\frac{-x^{10}}{2(x^5+x^3+1)^{2}}+c\)

substitute \(\x=\frac{1}{t})

Question 18

If \(f\) and \(g\) be continuous functions on \([0,a]\) such that \(f(x)=f(a-x)\) and \(g(x)+g(a-x)=4,\) then \(\int_{0}^{a} f(x)g(x) dx \) is equal to

A\(4\int_{0}^{a} f(x)dx \) B\(4\int_{0}^{a} f(x)dx \) C\(2\int_{0}^{a} f(x)dx \) D\(-3\int_{0}^{a} f(x)dx \)

Apply King's Theorem

Question 19

The value of \(\int_{-\pi/2}^{\pi/2}\frac{x^2 \cos x}{1+e^x}dx\) is equal to

A\(\frac{\pi^2}{4}-2\) B\(\frac{\pi^2}{4}+2\) C\(\pi^2-e^{-\pi/2}\) D\(\pi^2-e^{-\pi/2}\)

Use \(\int_{a}^{b} f(x) dx = \) \(\int_{a}^{b} f(a+b-x) dx\) then add the resulting integral with the given one and then apply integration by parts

Question 20

The value of \(\int_{\sqrt{\log 2}}^{\sqrt{\log 3} }\frac{x \sin x^2}{\sin x^2+\sin (\log 6 -x^2)}dx\) is equal to

A\(\frac{1}{4} \log \frac{3}{2}\) B\(\frac{1}{2} \log \frac{3}{2}\) C\(\log \frac{3}{2}\) D\(\frac{1}{6} \log \frac{3}{2}\)

put \(\log 6 - x^2 = t\) and then apply \(\int_{a}^{b} f(x) dx = \int_{a}^{0b} f(a+b-x) dx\)

Question 21

The value of \(\int_{-2}^{0} (x^3+3x^2+3x+3+(x+1) \cos (x+1)) dx\) is equal to

A\( 0 \) B\( 3 \) C\( 4 \) D\( 1 \)

Use \((x+1)^3\) to reduce the expression then substitute \(x+1=z\)

Question 22

Let \( f(x) = 7\tan^8 x + 7 \tan^6 x -3 \tan^4 x -3 \tan^2 x \), for all \(x \in \big(\frac{-\pi}{2},\frac{\pi}{2} \big)\). Then, the correct expression is

A \(\int_{0}^{\pi/4} x f(x) dx = \frac{1}{12}\) B \(\int_{0}^{\pi/4} x f(x) dx = \frac{1}{6}\) C \(\int_{0}^{\pi/4} x f(x) dx = \frac{5}{12}\) D \(\int_{0}^{\pi/4} x f(x) dx = \frac{3}{4}\)

Easy!

Question 23

The value of \(\int_{0}^{1/2} \frac{1+\sqrt{3}}{((x+1)^2(1-x)^6)^{1/4}} dx\) is equal to

A\( 2 \) B\( \frac{1}{\sqrt{3}} \) C\( 3 \) D\( 4 \)

After simplifying the integrand substitute \(x= \cos \theta\)

Question 24

The value of \(\,(5050)\frac{\int_{0}^{1} (1 - x^{50})^{100} \, dx}{\int_{0}^{1} (1 - x^{50})^{101} \, dx}\) is

A\( 5050 \) B\( 5051 \) C\( 1 \) D\( 5500 \)

\(If I_n \int_{0}^{1} (1-x^{50})^n dx\) then by use of integration by parts it is easy to see thet \((1+50n)I_n=50nI_{n-1}\). Take \((1-x^{50})^n\) as the first function

Question 25

The value of \(\int_{-2}^{2} |1-x^2| dx\) is equal to

A\(2\) B\(3\) C\(\frac{3}{4}\) D\(4\)

\(\int_{-2}^{2} |1-x^2| dx=\)\(\int_{-2}^{-1} -(1-x^2) dx\) \(+\int_{-1}^{1} (1-x^2) dx\)\(+\int_{1}^{2} -(1-x^2) dx\)

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