PYQs Guides Notes Mocks Board Questions JEE Mains Maths Definite Integration 196 JEE Mains Maths previous year questions (PYQs) from 88 papers, 2021 to 2026, with answers and worked solutions.
About this chapter Most-asked subtopics: Evaluating by Substitution, Parts and Partial Fractions (49), Greatest Integer, Modulus and Max-Min Integrands (48), Integral Equations and Leibniz's Rule (28). Updated 25 July 2026. #1 · Evaluating by Substitution, Parts and Partial Fractions
The value of the integral ∫ π 6 π 3 ( 4 − csc 2 x cos 4 x ) d x \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left( \frac{4 -\csc^{2}x}{\cos^{4}x} \right)dx ∫ 6 π 3 π ( c o s 4 x 4 − c s c 2 x ) d x is : A 11 3 \frac{11}{\sqrt{3}} 3 11 B 16 3 \frac{16}{\sqrt{3}} 3 16 C 32 3 3 \frac{32}{3\sqrt{3}} 3 3 32 D 64 3 3 \frac{64}{3\sqrt{3}} 3 3 64 #2 · Evaluating by Substitution, Parts and Partial Fractions
The value of the integral ∫ 0 2 x ( x 2 + x + 1 ) ( x + 1 ) ( x 4 + x 2 + 1 ) d x \int_{0}^{2} \frac{\sqrt{x\left( x^{2}+x+ 1 \right)}}{(\sqrt{x+ 1})\left( \sqrt{x^{4}+x^{2}+ 1} \right)}dx ∫ 0 2 ( x + 1 ) ( x 4 + x 2 + 1 ) x ( x 2 + x + 1 ) d x is equal to : A 1 3 log e ( 3 − 2 2 ) \frac{1}{3}\log_{e}(3 - 2\sqrt{2}) 3 1 log e ( 3 − 2 2 ) B 2 3 log e ( 4 + 2 ) \frac{2}{3}\log_{e}(4 + \sqrt{2}) 3 2 log e ( 4 + 2 ) C 2 3 log e ( 3 + 2 2 ) \frac{2}{3}\log_{e}(3 + 2\sqrt{2}) 3 2 log e ( 3 + 2 2 ) D 1 3 log e ( 1 + 6 2 ) \frac{1}{3}\log_{e}(1 + 6\sqrt{2}) 3 1 log e ( 1 + 6 2 ) #3 · Evaluating by Substitution, Parts and Partial Fractions
If ∫ π / 6 π / 4 ( cot ( x − π 3 ) cot ( x + π 3 ) + 1 ) d x = α log e ( 3 − 1 ) \int_{\pi/6}^{\pi/4} \left( \cot\left( x - \frac{\pi}{3} \right)\cot\left( x + \frac{\pi}{3} \right) + 1 \right)dx = \alpha\log_{e}(\sqrt{3} - 1) ∫ π /6 π /4 ( cot ( x − 3 π ) cot ( x + 3 π ) + 1 ) d x = α log e ( 3 − 1 ) , then 9 α 2 9\alpha^{2} 9 α 2 is equal to #4 · Evaluating by Substitution, Parts and Partial Fractions
Let ( 2 1 − a + 2 1 + a ) , f ( a ) , ( 3 a + 3 − a ) \left( 2^{1 - a}+2^{1 + a} \right),f(a),\left( 3^{a}+3^{- a} \right) ( 2 1 − a + 2 1 + a ) , f ( a ) , ( 3 a + 3 − a ) be a A. P. and α \alpha α be the minimum value of f ( a ) f(a) f ( a ) . Then the value of the integral ∫ log e ( α − 1 ) log e ( α ) d x ( e 2 x − e − 2 x ) \int_{\log_{e}(\alpha- 1)}^{\log_{e}(\alpha)} \frac{dx}{\left( e^{2x}-e^{- 2x} \right)} ∫ l o g e ( α − 1 ) l o g e ( α ) ( e 2 x − e − 2 x ) d x is : A 1 2 log e ( 4 3 ) \frac{1}{2}\log_{e}\left( \frac{4}{3} \right) 2 1 log e ( 3 4 ) B 1 4 log e ( 4 3 ) \frac{1}{4}\log_{e}\left( \frac{4}{3} \right) 4 1 log e ( 3 4 ) C 1 2 log e ( 8 5 ) \frac{1}{2}\log_{e}\left( \frac{8}{5} \right) 2 1 log e ( 5 8 ) D 1 4 log e ( 8 5 ) \frac{1}{4}\log_{e}\left( \frac{8}{5} \right) 4 1 log e ( 5 8 ) JEE Mains · 2026 · 4 Apr 2026 Shift 2 · Q70 Moderate Add #5 · Evaluating by Substitution, Parts and Partial Fractions
The integral ∫ 0 1 cot − 1 ( 1 + x + x 2 ) d x \int_{0}^{1} \cot^{- 1}\left( 1 + x +x^{2} \right)dx ∫ 0 1 cot − 1 ( 1 + x + x 2 ) d x is equal to : A 2 tan − 1 2 + 1 2 log e ( 5 4 ) + π 2 2\tan^{- 1}2 + \frac{1}{2}\log_{e}\left( \frac{5}{4} \right) + \frac{\pi}{2} 2 tan − 1 2 + 2 1 log e ( 4 5 ) + 2 π B 2 tan − 1 2 + 1 2 log e ( 5 4 ) − π 2 2\tan^{- 1}2 + \frac{1}{2}\log_{e}\left( \frac{5}{4} \right) - \frac{\pi}{2} 2 tan − 1 2 + 2 1 log e ( 4 5 ) − 2 π C 2 tan − 1 2 − 1 2 log e ( 5 4 ) + π 2 2\tan^{- 1}2 - \frac{1}{2}\log_{e}\left( \frac{5}{4} \right) + \frac{\pi}{2} 2 tan − 1 2 − 2 1 log e ( 4 5 ) + 2 π D 2 tan − 1 2 − 1 2 log e ( 5 4 ) − π 2 2\tan^{- 1}2 - \frac{1}{2}\log_{e}\left( \frac{5}{4} \right) - \frac{\pi}{2} 2 tan − 1 2 − 2 1 log e ( 4 5 ) − 2 π Done a few? Here’s your next move
JEE Mains · 2026 · 2 Apr 2026 Shift 1 · Q75 Moderate Add #6 · Evaluating by Substitution, Parts and Partial Fractions
If α = ∫ 0 2 3 log 2 ( x 2 + 4 ) d x + ∫ 2 4 2 x − 4 d x \alpha = \int_{0}^{2\sqrt{3}} \log_{2}\left( x^{2} + 4 \right)dx + \int_{2}^{4} \sqrt{2^{x} - 4}dx α = ∫ 0 2 3 log 2 ( x 2 + 4 ) d x + ∫ 2 4 2 x − 4 d x , then α 2 \alpha^{2} α 2 is equal to _ _ _ _ \_\_\_\_ ____ . #7 · Evaluating by Substitution, Parts and Partial Fractions
Let f f f be a polynomial function such that f ( x 2 + 1 ) = x 4 + 5 x 2 + 2 f\left( x^{2}+ 1 \right)=x^{4}+ 5x^{2}+ 2 f ( x 2 + 1 ) = x 4 + 5 x 2 + 2 , for all x ∈ R x\in\mathbb{R} x ∈ R . Then ∫ 0 3 f ( x ) d x \int_{0}^{3} f(x)dx ∫ 0 3 f ( x ) d x is equal to #8 · Evaluating by Substitution, Parts and Partial Fractions
The number of elements in the set S = { x : x ∈ [ 0 , 100 ] S = \left\{ x:x \in \lbrack 0,100\rbrack \right. S = { x : x ∈ [ 0 , 100 ] and ∫ 0 x t 2 sin ( x − t ) d t = x 2 } \left. \ \int_{0}^{x} t^{2}\sin(x - t)dt = x^{2} \right\} ∫ 0 x t 2 sin ( x − t ) d t = x 2 } is..... #9 · Evaluating by Substitution, Parts and Partial Fractions
If ∫ 0 1 4 cot − 1 ( 1 − 2 x + 4 x 2 ) d x = a tan − 1 ( 2 ) − b l o g e ( 5 ) \int_{0}^{1} 4\cot^{- 1}\left( 1 - 2x + 4x^{2} \right)dx = a\tan^{- 1}(2) - {blog}_{e}(5) ∫ 0 1 4 cot − 1 ( 1 − 2 x + 4 x 2 ) d x = a tan − 1 ( 2 ) − b l o g e ( 5 ) , where a , b ∈ N a,b \in N a , b ∈ N , then ( 2 a + b ) (2a + b) ( 2 a + b ) is equal to _ _ _ _ \_\_\_\_ ____ . JEE Mains · 2024 · 31 January 2024 · Q81 Moderate Add #10 · Evaluating by Substitution, Parts and Partial Fractions
If the integral 525 ∫ 0 π 2 sin 2 x cos 11 2 x ( 1 + cos 5 2 x ) 1 2 d x 525\int_{0}^{\frac{\pi}{2}} \sin2x\cos^{\frac{11}{2}}x\left( 1 +\cos^{\frac{5}{2}}x \right)^{\frac{1}{2}}dx 525 ∫ 0 2 π sin 2 x cos 2 11 x ( 1 + cos 2 5 x ) 2 1 d x is equal to ( n 2 − 64 ) (n\sqrt{2}- 64) ( n 2 − 64 ) , then n n n is equal to JEE Mains · 2023 · 11 April 2023 · Q61 Moderate Add #11 · Evaluating by Substitution, Parts and Partial Fractions
The value of the integral ∫ − log e 2 log e 2 e x ( log e ( e x + 1 + e 2 x ) ) d x \int_{-\log_{e}2}^{\log_{e}2} e^{x}\left( \log_{e}\left( e^{x}+\sqrt{1 +e^{2x}} \right) \right)dx ∫ − l o g e 2 l o g e 2 e x ( log e ( e x + 1 + e 2 x ) ) d x is equal to A log e ( 2 ( 2 + 5 ) 1 + 5 ) − 5 2 \log_{e}\left( \frac{2(2 + \sqrt{5})}{\sqrt{1 + \sqrt{5}}} \right) - \frac{\sqrt{5}}{2} log e ( 1 + 5 2 ( 2 + 5 ) ) − 2 5 B log e ( 2 ( 3 − 5 ) 2 1 + 5 ) + 5 2 \log_{e}\left( \frac{\sqrt{2}(3 - \sqrt{5})^{2}}{\sqrt{1 + \sqrt{5}}} \right) + \frac{\sqrt{5}}{2} log e ( 1 + 5 2 ( 3 − 5 ) 2 ) + 2 5 C log e ( ( 2 + 5 ) 2 1 + 5 ) + 5 2 \log_{e}\left( \frac{(2 + \sqrt{5})^{2}}{\sqrt{1 + \sqrt{5}}} \right) + \frac{\sqrt{5}}{2} log e ( 1 + 5 ( 2 + 5 ) 2 ) + 2 5 D log e ( 2 ( 2 + 5 ) 2 1 + 5 ) − 5 2 \log_{e}\left( \frac{\sqrt{2}(2 + \sqrt{5})^{2}}{\sqrt{1 + \sqrt{5}}} \right) - \frac{\sqrt{5}}{2} log e ( 1 + 5 2 ( 2 + 5 ) 2 ) − 2 5 JEE Mains · 2023 · 11 April 2023 · Q83 Moderate Add #12 · Evaluating by Substitution, Parts and Partial Fractions
For m , n > 0 m,n > 0 m , n > 0 , let α ( m , n ) = ∫ 0 2 t m ( 1 + 3 t ) n d t \alpha(m,n) =\int_{0}^{2} t^{m}(1 + 3t)^{n}dt α ( m , n ) = ∫ 0 2 t m ( 1 + 3 t ) n d t . If 11 α ( 10 , 6 ) + 18 α ( 11 , 5 ) = p ( 14 ) 6 11\alpha(10,6) + 18\alpha(11,5) = p(14)^{6} 11 α ( 10 , 6 ) + 18 α ( 11 , 5 ) = p ( 14 ) 6 , then p p p is equal to JEE Mains · 2025 · 4 April 2025 · Q143 Moderate Add #13 · Evaluating by Substitution, Parts and Partial Fractions
Let f ( x ) + 2 f ( 1 x ) = x 2 + 5 f(x) + 2f\left( \frac{1}{x} \right)=x^{2}+ 5 f ( x ) + 2 f ( x 1 ) = x 2 + 5 and 2 g ( x ) − 3 g ( 1 2 ) = x , x > 0 2g(x) - 3g\left( \frac{1}{2} \right)=x,x> 0 2 g ( x ) − 3 g ( 2 1 ) = x , x > 0 . If α = ∫ 1 2 f ( x ) d x \alpha=\int_{1}^{2} f(x)dx α = ∫ 1 2 f ( x ) d x , and β = ∫ 1 2 g ( x ) d x \beta=\int_{1}^{2} g(x)dx β = ∫ 1 2 g ( x ) d x , then the value of 9 α + β 9\alpha+\beta 9 α + β is : JEE Mains · 2023 · 31 January 2023 · Q70 Moderate Add #14 · Evaluating by Substitution, Parts and Partial Fractions
The value of ∫ π 3 π 2 ( 2 + 3 sin x ) sin x ( 1 + cos x ) d x \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2 + 3\sin x)}{\sin x(1 + \cos x)}dx ∫ 3 π 2 π s i n x ( 1 + c o s x ) ( 2 + 3 s i n x ) d x is equal to A 10 3 − 3 − log e 3 \frac{10}{3} - \sqrt{3} - \log_{e}\sqrt{3} 3 10 − 3 − log e 3 B 7 2 − 3 − log e 3 \frac{7}{2} - \sqrt{3} - \log_{e}\sqrt{3} 2 7 − 3 − log e 3 C − 2 + 3 3 + log e 3 - 2 + 3\sqrt{3} + \log_{e}\sqrt{3} − 2 + 3 3 + log e 3 D 10 3 − 3 + log e 3 \frac{10}{3} - \sqrt{3} + \log_{e}\sqrt{3} 3 10 − 3 + log e 3 JEE Mains · 2023 · 31 January 2023 · Q80 Moderate Add #15 · Evaluating by Substitution, Parts and Partial Fractions
Let α ∈ ( 0 , 1 ) \alpha \in (0,1) α ∈ ( 0 , 1 ) and β = log e ( 1 − α ) \beta =\log_{e}(1 - \alpha) β = log e ( 1 − α ) . Let P n ( x ) = x + x 2 2 + x 3 3 + ⋯ + x n n , x ∈ ( 0 , 1 ) P_{n}(x) = x +\frac{x^{2}}{2}+\frac{x^{3}}{3}+ \cdots +\frac{x^{n}}{n},x \in (0,1) P n ( x ) = x + 2 x 2 + 3 x 3 + ⋯ + n x n , x ∈ ( 0 , 1 ) . Then the integral ∫ 0 α t 50 1 − t d t \int_{0}^{\alpha} \frac{t^{50}}{1 - t}dt ∫ 0 α 1 − t t 50 d t is equal to A β + P 50 ( α ) \beta + P_{50}(\alpha) β + P 50 ( α ) B P 50 ( α ) − β P_{50}(\alpha) - \beta P 50 ( α ) − β C β − P 50 ( α ) \beta - P_{50}(\alpha) β − P 50 ( α ) D − ( β + P 50 ( α ) ) - \left( \beta + P_{50}(\alpha) \right) − ( β + P 50 ( α ) ) JEE Mains · 2023 · 31 January 2023 · Q153 Moderate Add #16 · Evaluating by Substitution, Parts and Partial Fractions
Let α > 0 \alpha > 0 α > 0 . If ∫ 0 α x x + α − x d x = 16 + 20 2 15 \int_{0}^{\alpha} \frac{x}{\sqrt{x + \alpha}-\sqrt{x}}dx =\frac{16 + 20\sqrt{2}}{15} ∫ 0 α x + α − x x d x = 15 16 + 20 2 , then α \alpha α is equal to: JEE Mains · 2024 · 8 April 2024 · Q166 Moderate Add #17 · Evaluating by Substitution, Parts and Partial Fractions
Let ∫ α log e 4 d x e x − 1 = π 6 \int_{\alpha}^{\log_{e}4} \frac{dx}{\sqrt{e^{x}- 1}}=\frac{\pi}{6} ∫ α l o g e 4 e x − 1 d x = 6 π . Then e α e^{\alpha} e α and e − α e^{-\alpha} e − α are the roots of the equation : A 2 x 2 − 5 x + 2 = 0 2x^{2} - 5x + 2 = 0 2 x 2 − 5 x + 2 = 0 B x 2 − 2 x − 8 = 0 x^{2} - 2x - 8 = 0 x 2 − 2 x − 8 = 0 C 2 x 2 − 5 x − 2 = 0 2x^{2} - 5x - 2 = 0 2 x 2 − 5 x − 2 = 0 D x 2 + 2 x − 8 = 0 x^{2} + 2x - 8 = 0 x 2 + 2 x − 8 = 0 JEE Mains · 2024 · 6 April 2024 · Q63 Moderate Add #18 · Evaluating by Substitution, Parts and Partial Fractions
∫ 0 π / 4 cos 2 x sin 2 x ( cos 3 x + sin 3 x ) 2 d x \int_{0}^{\pi/4} \frac{\cos^{2}x\sin^{2}x}{\left( \cos^{3}x+\sin^{3}x \right)^{2}}dx ∫ 0 π /4 ( c o s 3 x + s i n 3 x ) 2 c o s 2 x s i n 2 x d x is equal to JEE Mains · 2024 · 9 April 2024 · Q165 Moderate Add #19 · Evaluating by Substitution, Parts and Partial Fractions
The integral ∫ 1 / 4 3 / 4 cos ( 2 cot − 1 1 − x 1 + x ) d x \int_{1/4}^{3/4} \cos\left( 2\cot^{- 1}\sqrt{\frac{1 -x}{1 +x}} \right)dx ∫ 1/4 3/4 cos ( 2 cot − 1 1 + x 1 − x ) d x is equal to: JEE Mains · 2024 · 9 April 2024 · Q168 Moderate Add #20 · Evaluating by Substitution, Parts and Partial Fractions
The value of the integral ∫ − 1 2 log e ( x + x 2 + 1 ) d x \int_{- 1}^{2} \log_{e}\left( x+\sqrt{x^{2}+ 1} \right)dx ∫ − 1 2 log e ( x + x 2 + 1 ) d x is: A 5 − 2 + log e ( 9 + 4 5 1 + 2 ) \sqrt{5} - \sqrt{2} + \log_{e}\left( \frac{9 + 4\sqrt{5}}{1 + \sqrt{2}} \right) 5 − 2 + log e ( 1 + 2 9 + 4 5 ) B 2 − 5 + log e ( 9 + 4 5 1 + 2 ) \sqrt{2} - \sqrt{5} + \log_{e}\left( \frac{9 + 4\sqrt{5}}{1 + \sqrt{2}} \right) 2 − 5 + log e ( 1 + 2 9 + 4 5 ) C 5 − 2 + log e ( 7 + 4 5 1 + 2 ) \sqrt{5} - \sqrt{2} + \log_{e}\left( \frac{7 + 4\sqrt{5}}{1 + \sqrt{2}} \right) 5 − 2 + log e ( 1 + 2 7 + 4 5 ) D 2 − 5 + log e ( 7 + 4 5 1 + 2 ) \sqrt{2} - \sqrt{5} + \log_{e}\left( \frac{7 + 4\sqrt{5}}{1 + \sqrt{2}} \right) 2 − 5 + log e ( 1 + 2 7 + 4 5 ) JEE Mains · 2023 · 1 February 2023 · Q88 Moderate Add #21 · Evaluating by Substitution, Parts and Partial Fractions
If ∫ 0 1 ( x 21 + x 14 + x 7 ) ( 2 x 14 + 3 x 7 + 6 ) 1 / 7 d x = 1 l ( 11 ) m / n \int_{0}^{1} \left( x^{21}+x^{14}+x^{7} \right)\left( 2x^{14}+ 3x^{7}+ 6 \right)^{1/7}dx =\frac{1}{l}(11)^{m/n} ∫ 0 1 ( x 21 + x 14 + x 7 ) ( 2 x 14 + 3 x 7 + 6 ) 1/7 d x = l 1 ( 11 ) m / n where l , m , n ∈ N , m l,m,n \in N,m l , m , n ∈ N , m and n n n are coprime then l + m + n l + m + n l + m + n is equal to JEE Mains · 2024 · 4 April 2024 · Q90 Moderate Add #22 · Evaluating by Substitution, Parts and Partial Fractions
If ∫ 0 π 4 sin 2 x 1 + sin x c o s x d x = 1 a log e ( a 3 ) + π b 3 \int_{0}^{\frac{\pi}{4}} \frac{\sin^{2}x}{1 + \sin xcosx}dx =\frac{1}{a}\log_{e}\left( \frac{a}{3} \right)+\frac{\pi}{b\sqrt{3}} ∫ 0 4 π 1 + s i n x cos x s i n 2 x d x = a 1 log e ( 3 a ) + b 3 π , where a a a , b ∈ N b \in N b ∈ N , then a + b a + b a + b is equal to JEE Mains · 2023 · 6 April 2023 · Q61 Moderate Add #23 · Evaluating by Substitution, Parts and Partial Fractions
Let 5 f ( x ) + 4 f ( 1 x ) = 1 x + 3 , x > 0 5f(x) + 4f\left( \frac{1}{x} \right)=\frac{1}{x}+ 3,x > 0 5 f ( x ) + 4 f ( x 1 ) = x 1 + 3 , x > 0 . Then 18 ∫ 1 2 f ( x ) d x 18\int_{1}^{2} f(x)dx 18 ∫ 1 2 f ( x ) d x is equal to: A 10 log e 2 − 6 10\log_{e}2 - 6 10 log e 2 − 6 B 10 log e 2 + 6 10\log_{e}2 + 6 10 log e 2 + 6 C 5 log e 2 + 3 5\log_{e}2 + 3 5 log e 2 + 3 D 5 log e 2 − 3 5\log_{e}2 - 3 5 log e 2 − 3 JEE Mains · 2024 · 1 February 2024 · Q161 Moderate Add #24 · Evaluating by Substitution, Parts and Partial Fractions
If ∫ 0 π 3 cos 4 x d x = a π + b 3 \int_{0}^{\frac{\pi}{3}} \cos^{4}xdx = a\pi + b\sqrt{3} ∫ 0 3 π cos 4 x d x = aπ + b 3 , where a a a and b b b are rational numbers, then 9 a + 8 b 9a + 8\text{ }b 9 a + 8 b is equal to : JEE Mains · 2022 · 29 July 2022 · Q66 Moderate Add #25 · Evaluating by Substitution, Parts and Partial Fractions
The integral ∫ 0 π 2 1 3 + 2 sin x + cos x d x \int_{0}^{\frac{\pi}{2}} \frac{1}{3 + 2\sin x + \cos x}dx ∫ 0 2 π 3 + 2 s i n x + c o s x 1 d x is equal to: A tan − 1 ( 2 ) \tan^{- 1}(2) tan − 1 ( 2 ) B tan − 1 ( 2 ) − π 4 \tan^{- 1}(2) - \frac{\pi}{4} tan − 1 ( 2 ) − 4 π C 1 2 tan − 1 ( 2 ) − π 8 \frac{1}{2}\tan^{- 1}(2) - \frac{\pi}{8} 2 1 tan − 1 ( 2 ) − 8 π D JEE Mains · 2022 · 27 July 2022 · Q68 Moderate Add #26 · Evaluating by Substitution, Parts and Partial Fractions
I = ∫ π / 4 π / 3 ( 8 sin x − sin 2 x x ) d x \ I =\int_{\pi/4}^{\pi/3} \left( \frac{8\sin x - \sin2x}{x} \right)dx I = ∫ π /4 π /3 ( x 8 s i n x − s i n 2 x ) d x . Then A π 2 < I < 3 π 4 \frac{\pi}{2} < I < \frac{3\pi}{4} 2 π < I < 4 3 π B π 5 < I < 5 π 12 \frac{\pi}{5} < I < \frac{5\pi}{12} 5 π < I < 12 5 π C 5 π 12 < I < 2 3 π \frac{5\pi}{12} < I < \frac{\sqrt{2}}{3}\pi 12 5 π < I < 3 2 π D 3 π 4 < I < π \frac{3\pi}{4} < I < \pi 4 3 π < I < π JEE Mains · 2022 · 25 June 2022 · Q173 Moderate Add #27 · Evaluating by Substitution, Parts and Partial Fractions
The value of b > 3 b> 3 b > 3 for which 12 ∫ 3 b 1 ( x 2 − 1 ) ( x 2 − 4 ) d x = log e ( 49 40 ) 12\int_{3}^{b} \frac{1}{\left( x^{2}- 1 \right)\left( x^{2}- 4 \right)}dx =\log_{e}\left( \frac{49}{40} \right) 12 ∫ 3 b ( x 2 − 1 ) ( x 2 − 4 ) 1 d x = log e ( 40 49 ) , is equal to JEE Mains · 2023 · 29 January 2023 · Q152 Moderate Add #28 · Evaluating by Substitution, Parts and Partial Fractions
The value of the integral ∫ 1 2 ( t 4 + 1 t 6 + 1 ) d t \int_{1}^{2} \left( \frac{t^{4}+ 1}{t^{6}+ 1} \right)dt ∫ 1 2 ( t 6 + 1 t 4 + 1 ) d t is A tan − 1 2 − 1 3 tan − 1 8 + π 3 \tan^{- 1}2 - \frac{1}{3}\tan^{- 1}8 + \frac{\pi}{3} tan − 1 2 − 3 1 tan − 1 8 + 3 π B tan − 1 1 2 + 1 3 tan − 1 8 − π 3 \tan^{- 1}\frac{1}{2} + \frac{1}{3}\tan^{- 1}8 - \frac{\pi}{3} tan − 1 2 1 + 3 1 tan − 1 8 − 3 π C tan − 1 1 2 − 1 3 tan − 1 8 + π 3 \tan^{- 1}\frac{1}{2} - \frac{1}{3}\tan^{- 1}8 + \frac{\pi}{3} tan − 1 2 1 − 3 1 tan − 1 8 + 3 π D tan − 1 2 + 1 3 tan − 1 8 − π 3 \tan^{- 1}2 + \frac{1}{3}\tan^{- 1}8 - \frac{\pi}{3} tan − 1 2 + 3 1 tan − 1 8 − 3 π JEE Mains · 2023 · 13 April 2023 · Q61 Moderate Add #29 · Evaluating by Substitution, Parts and Partial Fractions
∫ 0 ∞ 6 e 3 x + 6 e 2 x + 11 e x + 6 d x \int_{0}^{\infty} \frac{6}{e^{3x}+ 6e^{2x}+ 11e^{x}+ 6}dx ∫ 0 ∞ e 3 x + 6 e 2 x + 11 e x + 6 6 d x A log e ( 512 81 ) \log_{e}\left( \frac{512}{81} \right) log e ( 81 512 ) B log e ( 32 27 ) \log_{e}\left( \frac{32}{27} \right) log e ( 27 32 ) C log e ( 256 81 ) \log_{e}\left( \frac{256}{81} \right) log e ( 81 256 ) D log e ( 64 27 ) \log_{e}\left( \frac{64}{27} \right) log e ( 27 64 ) JEE Mains · 2023 · 13 April 2023 · Q179 Moderate Add #30 · Evaluating by Substitution, Parts and Partial Fractions
Let f n = ∫ 0 π 2 ( ∑ k = 1 n sin k − 1 x ) ( ∑ k = 1 n ( 2 k − 1 ) sin k − 1 x ) cos x f_{n}=\int_{0}^{\frac{\pi}{2}} \left( \sum_{k = 1}^{n} \sin^{k - 1}x \right)\left( \sum_{k = 1}^{n} (2k - 1)\sin^{k - 1}x \right)\cos x f n = ∫ 0 2 π ( ∑ k = 1 n sin k − 1 x ) ( ∑ k = 1 n ( 2 k − 1 ) sin k − 1 x ) cos x d x , n ∈ N dx,n \in N d x , n ∈ N . Then f 21 − f 20 f_{21}-f_{20} f 21 − f 20 is equal to JEE Mains · 2022 · 28 July 2022 · Q84 Moderate Add #31 · Evaluating by Substitution, Parts and Partial Fractions
If ∫ 0 3 15 x 3 1 + x 2 + ( 1 + x 2 ) 3 d x = α 2 + β 3 \int_{0}^{\sqrt{3}} \frac{15x^{3}}{\sqrt{1 +x^{2}+\sqrt{\left( 1 +x^{2} \right)^{3}}}}dx = \alpha\sqrt{2}+ \beta\sqrt{3} ∫ 0 3 1 + x 2 + ( 1 + x 2 ) 3 15 x 3 d x = α 2 + β 3 , where α , β \alpha,\beta α , β are integers, then α + β \alpha + \beta α + β is equal to JEE Mains · 2022 · 28 July 2022 · Q180 Moderate Add #32 · Evaluating by Substitution, Parts and Partial Fractions
The value of the integral ∫ 0 π 2 60 sin ( 6 x ) sin x d x \int_{0}^{\frac{\pi}{2}} 60\frac{\sin(6x)}{\sin x}dx ∫ 0 2 π 60 s i n x s i n ( 6 x ) d x is equal to JEE Mains · 2022 · 25 July 2022 · Q177 Moderate Add #33 · Evaluating by Substitution, Parts and Partial Fractions
Let a n = ∫ − 1 n ( 1 + x 2 + x 2 3 + x 3 4 + … … . + x n − 1 n ) d x a_{n}=\int_{- 1}^{n} \left( 1 +\frac{x}{2}+\frac{x^{2}}{3}+\frac{x^{3}}{4}+ \ldots\ldots. +\frac{x^{n - 1}}{n} \right)dx a n = ∫ − 1 n ( 1 + 2 x + 3 x 2 + 4 x 3 + …… . + n x n − 1 ) d x for n ∈ N n \in N n ∈ N . Then the sum of all the elements of the set { n ∈ N : a n ∈ ( 2 , 30 ) } \left\{ n \in N:a_{n}\in (2,30) \right\} { n ∈ N : a n ∈ ( 2 , 30 ) } is JEE Mains · 2022 · 26 June 2022 · Q177 Moderate Add #34 · Evaluating by Substitution, Parts and Partial Fractions
The integral 24 π ∫ 0 2 ( 2 − x 2 ) d x ( 2 + x 2 ) 4 + x 4 \frac{24}{\pi}\int_{0}^{\sqrt{2}} \frac{\left( 2 -x^{2} \right)dx}{\left( 2 +x^{2} \right)\sqrt{4 +x^{4}}} π 24 ∫ 0 2 ( 2 + x 2 ) 4 + x 4 ( 2 − x 2 ) d x is equal to #35 · Evaluating by Substitution, Parts and Partial Fractions
The integral 80 ∫ 0 π 4 ( sin θ + cos θ 9 + 16 sin 2 θ ) d θ 80\int_{0}^{\frac{\pi}{4}} \left( \frac{\sin\theta + \cos\theta}{9 + 16\sin2\theta} \right)d\theta 80 ∫ 0 4 π ( 9 + 16 s i n 2 θ s i n θ + c o s θ ) d θ is equal to : #36 · Evaluating by Substitution, Parts and Partial Fractions
4 ∫ 0 1 ( 1 3 + x 2 + 1 + x 2 ) d x − 3 log e ( 3 ) 4\int_{0}^{1} \left( \frac{1}{\sqrt{3 +x^{2}}+\sqrt{1 +x^{2}}} \right)dx - 3\log_{e}(\sqrt{3}) 4 ∫ 0 1 ( 3 + x 2 + 1 + x 2 1 ) d x − 3 log e ( 3 ) is equal to: A 2 + 2 + log e ( 1 + 2 ) 2 + \sqrt{2} + \log_{e}(1 + \sqrt{2}) 2 + 2 + log e ( 1 + 2 ) B 2 − 2 − log e ( 1 + 2 ) 2 - \sqrt{2} - \log_{e}(1 + \sqrt{2}) 2 − 2 − log e ( 1 + 2 ) C 2 + 2 − log e ( 1 + 2 ) 2 + \sqrt{2} - \log_{e}(1 + \sqrt{2}) 2 + 2 − log e ( 1 + 2 ) D 2 − 2 + log e ( 1 + 2 ) 2 - \sqrt{2} + \log_{e}(1 + \sqrt{2}) 2 − 2 + log e ( 1 + 2 ) JEE Mains · 2025 · 28 Jan 2025 · Q137 Moderate Add #37 · Evaluating by Substitution, Parts and Partial Fractions
Let f : R → R f:R\rightarrow R f : R → R be a twice differentiable function such that f ( 2 ) = 1 f(2) = 1 f ( 2 ) = 1 . If F ( x ) = x f ( x ) F(x) = xf(x) F ( x ) = x f ( x ) for all x ∈ R x \in R x ∈ R , ∫ 0 2 x F ′ ( x ) d x = 6 \int_{0}^{2} xF^{'}(x)dx= 6 ∫ 0 2 x F ′ ( x ) d x = 6 and ∫ 0 2 x 2 F ′ ′ ( x ) d x = 40 \int_{0}^{2} x^{2}F^{''}(x)dx= 40 ∫ 0 2 x 2 F ′′ ( x ) d x = 40 , then F ′ ( 2 ) + ∫ 0 2 F ( x ) d x F^{'}(2) +\int_{0}^{2} F(x)dx F ′ ( 2 ) + ∫ 0 2 F ( x ) d x is equal to : JEE Mains · 2025 · 22 Jan 2025 · Q65 Moderate Add #38 · Evaluating by Substitution, Parts and Partial Fractions
Let for f ( x ) = 7 tan 8 x + 7 tan 6 x − 3 tan 4 x − 3 tan 2 x f(x) = 7\tan^{8}x + 7\tan^{6}x - 3\tan^{4}x - 3\tan^{2}x f ( x ) = 7 tan 8 x + 7 tan 6 x − 3 tan 4 x − 3 tan 2 x , I 1 = ∫ 0 π / 4 f ( x ) d x I_{1}=\int_{0}^{\pi/4} f(x)dx I 1 = ∫ 0 π /4 f ( x ) d x and I 2 = ∫ 0 π / 4 x f ( x ) d x I_{2}=\int_{0}^{\pi/4} xf(x)dx I 2 = ∫ 0 π /4 x f ( x ) d x . Then 7 I 1 + 12 I 2 7I_{1}+ 12I_{2} 7 I 1 + 12 I 2 is equal to : JEE Mains · 2023 · 25 Jan 2023 · Q160 Moderate Add #39 · Evaluating by Substitution, Parts and Partial Fractions
The integral 16 ∫ 1 2 d x x 3 ( x 2 + 2 ) 2 16\int_{1}^{2} \frac{dx}{x^{3}\left(x^{2}+2\right)^{2}} 16 ∫ 1 2 x 3 ( x 2 + 2 ) 2 d x is equal to A 11 12 − log e 4 \frac{11}{12}-\log_{e}4 12 11 − log e 4 B 11 6 − log e 4 \frac{11}{6}-\log_{e}4 6 11 − log e 4 C 11 6 + log e 4 \frac{11}{6}+\log_{e}4 6 11 + log e 4 D 11 12 + log e 4 \frac{11}{12}+\log_{e}4 12 11 + log e 4 JEE Mains · 2023 · 24 Jan 2023 · Q159 Moderate Add #40 · Evaluating by Substitution, Parts and Partial Fractions
∫ 3 2 4 3 3 4 48 9 − 4 x 2 d x \ \int_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}} \frac{48}{\sqrt{9 - 4x^{2}}}dx ∫ 4 3 2 4 3 3 9 − 4 x 2 48 d x is equal to JEE Mains · 2024 · 30 Jan 2024 · Q162 Moderate Add #41 · Evaluating by Substitution, Parts and Partial Fractions
Let f : R → R f:R \rightarrow R f : R → R be defined f ( x ) = a e 2 x + b e x + c x f(x) = ae^{2x}+ be^{x}+ cx f ( x ) = a e 2 x + b e x + c x . If f ( 0 ) = − 1 , f ′ ( log e 2 ) = 21 f(0) = - 1,f^{'}\left( \log_{e}2 \right)= 21 f ( 0 ) = − 1 , f ′ ( log e 2 ) = 21 and ∫ 0 log e 4 ( f ( x ) − c x ) d x = 39 2 \int_{0}^{\log_{e}4} (f(x) - cx)dx =\frac{39}{2} ∫ 0 l o g e 4 ( f ( x ) − c x ) d x = 2 39 , then the value of ∣ a + b + c ∣ |a + b + c| ∣ a + b + c ∣ equals : JEE Mains · 2024 · 30 Jan 2024 · Q169 Moderate Add #42 · Evaluating by Substitution, Parts and Partial Fractions
Let f : R → R f:R \rightarrow R f : R → R be a function defined f ( x ) = x ( 1 + x 4 ) 1 / 4 f(x) =\frac{x}{\left( 1 +x^{4} \right)^{1/4}} f ( x ) = ( 1 + x 4 ) 1/4 x and g ( x ) = f ( f ( f ( f ( x ) ) ) ) g(x) = f(f(f(f(x)))) g ( x ) = f ( f ( f ( f ( x )))) then 18 ∫ 0 2 5 x 2 g ( x ) d x 18\int_{0}^{\sqrt{2\sqrt{5}}} x^{2}g(x)dx 18 ∫ 0 2 5 x 2 g ( x ) d x JEE Mains · 2024 · 5 Apr 2024 · Q69 Moderate Add #43 · Evaluating by Substitution, Parts and Partial Fractions
The integral ∫ 0 π 4 136 sin x 3 sin x + 5 cos x d x \int_{0}^{\frac{\pi}{4}} \frac{136\sin x}{3\sin x + 5\cos x}dx ∫ 0 4 π 3 s i n x + 5 c o s x 136 s i n x d x is equal to: A 3 π − 50 log e 2 + 20 log e 5 3\pi - 50\log_{e}2 + 20\log_{e}5 3 π − 50 log e 2 + 20 log e 5 B 3 π − 25 log e 2 + 10 log e 5 3\pi - 25\log_{e}2 + 10\log_{e}5 3 π − 25 log e 2 + 10 log e 5 C 3 π − 10 log e ( 2 2 ) + 10 log e 5 3\pi - 10\log_{e}(2\sqrt{2}) + 10\log_{e}5 3 π − 10 log e ( 2 2 ) + 10 log e 5 D 3 π − 30 log e 2 + 20 log e 5 3\pi - 30\log_{e}2 + 20\log_{e}5 3 π − 30 log e 2 + 20 log e 5 JEE Mains · 2024 · 27 Jan 2024 · Q69 Moderate Add #44 · Evaluating by Substitution, Parts and Partial Fractions
If ∫ 0 1 1 3 + x + 1 + x d x = a + b 2 + c 3 \int_{0}^{1} \frac{1}{\sqrt{3 + x}+\sqrt{1 + x}}dx = a + b\sqrt{2}+ c\sqrt{3} ∫ 0 1 3 + x + 1 + x 1 d x = a + b 2 + c 3 , where a , b , c a,b,c a , b , c are rational numbers, then 2 a + 3 b − 4 c 2a + 3b - 4c 2 a + 3 b − 4 c is equal to : JEE Mains · 2021 · Paper 17 · Q85 Moderate Add #45 · Evaluating by Substitution, Parts and Partial Fractions
If ∫ 0 π ( sin 3 x ) e − sin 2 x d x = α − β e ∫ 0 1 t e t d t \int_{0}^{\pi} \left( \sin^{3}x \right)e^{-\sin^{2}x}dx=\alpha-\frac{\beta}{e}\int_{0}^{1} \sqrt{t}e^{t}dt ∫ 0 π ( sin 3 x ) e − s i n 2 x d x = α − e β ∫ 0 1 t e t d t , then α + β \alpha+\beta α + β is equal to JEE Mains · 2021 · Paper 21 · Q79 Moderate Add #46 · Evaluating by Substitution, Parts and Partial Fractions
The value of the integral ∫ 0 1 x d x ( 1 + x ) ( 1 + 3 x ) ( 3 + x ) \int_{0}^{1} \frac{\sqrt{x}dx}{(1 +x)(1 + 3x)(3 +x)} ∫ 0 1 ( 1 + x ) ( 1 + 3 x ) ( 3 + x ) x d x is: A π 8 ( 1 − 3 2 ) \frac{\pi}{8}\left( 1 - \frac{\sqrt{3}}{2} \right) 8 π ( 1 − 2 3 ) B π 4 ( 1 − 3 6 ) \frac{\pi}{4}\left( 1 - \frac{\sqrt{3}}{6} \right) 4 π ( 1 − 6 3 ) C π 8 ( 1 − 3 6 ) \frac{\pi}{8}\left( 1 - \frac{\sqrt{3}}{6} \right) 8 π ( 1 − 6 3 ) D π 4 ( 1 − 3 2 ) \frac{\pi}{4}\left( 1 - \frac{\sqrt{3}}{2} \right) 4 π ( 1 − 2 3 ) JEE Mains · 2021 · Paper 10 · Q68 Moderate Add #47 · Evaluating by Substitution, Parts and Partial Fractions
Let g ( x ) = ∫ 0 x f ( t ) d t g(x) =\int_{0}^{x} f(t)dt g ( x ) = ∫ 0 x f ( t ) d t , where f f f is continuous function in [ 0 , 3 ] \lbrack 0,3\rbrack [ 0 , 3 ] such that 1 3 ≤ f ( t ) ≤ 1 \frac{1}{3}\leq f(t) \leq 1 3 1 ≤ f ( t ) ≤ 1 for all t ∈ [ 0 , 1 ] t \in \lbrack 0,1\rbrack t ∈ [ 0 , 1 ] and 0 ≤ f ( t ) ≤ 1 2 0 \leq f(t) \leq\frac{1}{2} 0 ≤ f ( t ) ≤ 2 1 for all t ∈ ( 1 , 3 ] t \in (1,3\rbrack t ∈ ( 1 , 3 ] . The largest possible interval in which g ( 3 ) g(3) g ( 3 ) lies is: A [ − 1 , − 1 2 ] \left\lbrack - 1, - \frac{1}{2} \right\rbrack [ − 1 , − 2 1 ] B [ − 3 2 , − 1 ] \left\lbrack - \frac{3}{2}, - 1 \right\rbrack [ − 2 3 , − 1 ] C [ 1 3 , 2 ] \left\lbrack \frac{1}{3},2 \right\rbrack [ 3 1 , 2 ] D [ 1 , 3 ] \lbrack 1,3\rbrack [ 1 , 3 ] JEE Mains · 2021 · Paper 8 · Q61 Moderate Add #48 · Evaluating by Substitution, Parts and Partial Fractions
Let f : R → R f:R \rightarrow R f : R → R be defined as f ( x ) = e − x sin x f(x) =e^{- x}\sin x f ( x ) = e − x sin x . If F : [ 0 , 1 ] → R F:\lbrack 0,1\rbrack \rightarrow R F : [ 0 , 1 ] → R is a differentiable function such that F ( x ) = ∫ 0 x f ( t ) d t F(x) =\int_{0}^{x} f(t)dt F ( x ) = ∫ 0 x f ( t ) d t , then the value of ∫ 0 1 ( F ′ ( x ) + f ( x ) ) e x d x \int_{0}^{1} \left( F^{'}(x) +f(x) \right)e^{x}dx ∫ 0 1 ( F ′ ( x ) + f ( x ) ) e x d x lies in the interval A [ 327 360 , 329 360 ] \left\lbrack \frac{327}{360},\frac{329}{360} \right\rbrack [ 360 327 , 360 329 ] B [ 330 360 , 331 360 ] \left\lbrack \frac{330}{360},\frac{331}{360} \right\rbrack [ 360 330 , 360 331 ] C [ 331 360 , 334 360 ] \left\lbrack \frac{331}{360},\frac{334}{360} \right\rbrack [ 360 331 , 360 334 ] D [ 335 360 , 336 360 ] \left\lbrack \frac{335}{360},\frac{336}{360} \right\rbrack [ 360 335 , 360 336 ] JEE Mains · 2021 · Paper 6 · Q77 Moderate Add #49 · Evaluating by Substitution, Parts and Partial Fractions
Let P ( x ) = x 2 + b x + c P(x) =x^{2}+ bx + c P ( x ) = x 2 + b x + c be a quadratic polynomial with real coefficients such that ∫ 0 1 P ( x ) d x = 1 \int_{0}^{1} P(x)dx = 1 ∫ 0 1 P ( x ) d x = 1 and P ( x ) P(x) P ( x ) leaves remainder 5 when it is divided by ( x − 2 ) (x- 2) ( x − 2 ) . Then the value of 9 ( b + c ) 9(b+c) 9 ( b + c ) is equal to: #50 · The a + b - x Property and Other Symmetries
The value of the integral ∫ 0 ∞ log e ( x ) x 2 + 4 d x \int_{0}^{\infty} \frac{\log_{e}(x)}{x^{2}+ 4}dx ∫ 0 ∞ x 2 + 4 l o g e ( x ) d x is: A π log e ( 2 ) 2 \frac{\pi\log_{e}(2)}{2} 2 π l o g e ( 2 ) B π log e ( 2 ) 4 \frac{\pi\log_{e}(2)}{4} 4 π l o g e ( 2 ) C 1 + π log e ( 2 ) 1 + \pi\log_{e}(2) 1 + π log e ( 2 ) D 2 + π log e ( 2 ) 2 + \pi\log_{e}(2) 2 + π log e ( 2 ) Showing the first 50 of 196.
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