PYQs Guides Notes Mocks Board Questions MHT-CET Maths Definite Integration 132 MHT-CET Maths previous year questions (PYQs) from 68 papers, 2021 to 2026, with answers and worked solutions.
About this chapter Difficulty: 14 easy · 60 moderate · 58 hard. Most-asked subtopics: Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts (45), King's Property — f(a + b − x) and the f/(f + g) Family (28), Modulus and Greatest-Integer Integrands — Split the Interval (25). Updated 13 October 2026. #1 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
The value of lim n → ∞ ( n ( n + 1 ) 2 n + 1 + n ( n + 2 ) 2 ( 2 n + 2 ) + n ( n + 3 ) 3 ( 2 n + 3 ) + ⋅ s + 1 2 n 3 ) \displaystyle\lim_{n \to \infty}\left(\dfrac{n}{(n + 1)\sqrt{2n + 1}} + \dfrac{n}{(n + 2)\sqrt{2(2n + 2)}} + \dfrac{n}{(n + 3)\sqrt{3(2n + 3)}} + \cdot s + \dfrac{1}{2n\sqrt{3}}\right) n → ∞ lim ( ( n + 1 ) 2 n + 1 n + ( n + 2 ) 2 ( 2 n + 2 ) n + ( n + 3 ) 3 ( 2 n + 3 ) n + ⋅ s + 2 n 3 1 ) is: #2 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If ∫ 0 f ( x ) d t = x cos ( π x ) \displaystyle\int_0^{f(x)} dt = x\cos(\pi x) ∫ 0 f ( x ) d t = x cos ( π x ) then f ′ ( 4 ) = … f'(4) = \ldots f ′ ( 4 ) = … #3 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If f ( x ) = sin ( tan − 1 x ) f(x) = \sin(\tan^{-1}x) f ( x ) = sin ( tan − 1 x ) , then ∫ 0 1 x f ′ ′ ( x ) d x \int_0^1 x f''(x)dx ∫ 0 1 x f ′′ ( x ) d x is: A 3 2 \dfrac{\sqrt{3}}{2} 2 3 B − 1 2 2 -\dfrac{1}{2\sqrt{2}} − 2 2 1 C 1 2 \dfrac{1}{\sqrt{2}} 2 1 D 1 2 2 \dfrac{1}{2\sqrt{2}} 2 2 1 #4 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If ∫ − 1 4 f ( x ) d x = 9 \int_{-1}^{4} f(x)dx = 9 ∫ − 1 4 f ( x ) d x = 9 and ∫ 2 4 ( 7 − f ( x ) ) d x = 10 \int_{2}^{4}(7 - f(x))dx = 10 ∫ 2 4 ( 7 − f ( x )) d x = 10 , then ∫ 2 − 1 f ( x ) d x = \int_{2}^{-1} f(x)dx = ∫ 2 − 1 f ( x ) d x = #5 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
The value of ∫ 1 2 1 x ( 1 + log x ) 2 d x \int_1^2 \dfrac{1}{x(1 + \log x)^2}\,dx ∫ 1 2 x ( 1 + log x ) 2 1 d x is....... A log 2 1 − log 2 \dfrac{\log 2}{1 - \log 2} 1 − log 2 log 2 B log 2 1 + log 2 \dfrac{\log 2}{1 + \log 2} 1 + log 2 log 2 C log 2 − 1 + log 2 \dfrac{\log 2}{-1 + \log 2} − 1 + log 2 log 2 D − log 2 1 + log 2 \dfrac{-\log 2}{1 + \log 2} 1 + log 2 − log 2 Done a few? Here’s your next move
#6 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
The value of the integral ∫ 1 / 2 2 1 x sin 101 ( x − 1 x ) d x \int_{1/2}^{2}\dfrac{1}{x}\mathrm{\sin}^{101}\left(x - \dfrac{1}{x}\right) dx ∫ 1/2 2 x 1 sin 101 ( x − x 1 ) d x is equal to: #7 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
The value of I = ∫ 1 3 e x ( x + 2 ) 2 ( x + 4 ) 2 d x I = \int_1^3 e^x\dfrac{(x + 2)^2}{(x + 4)^2}\, dx I = ∫ 1 3 e x ( x + 4 ) 2 ( x + 2 ) 2 d x is: A e ( 3 e 2 7 + 1 5 ) e\left(\dfrac{3e^2}{7} + \dfrac{1}{5}\right) e ( 7 3 e 2 + 5 1 ) B e ( 3 e 2 7 − 1 5 ) e\left(\dfrac{3e^2}{7} - \dfrac{1}{5}\right) e ( 7 3 e 2 − 5 1 ) C e 2 ( 3 e 7 − 1 5 ) e^2\left(\dfrac{3e}{7} - \dfrac{1}{5}\right) e 2 ( 7 3 e − 5 1 ) D e ( 3 e 2 7 − e 5 ) e\left(\dfrac{3e^2}{7} - \dfrac{e}{5}\right) e ( 7 3 e 2 − 5 e ) #8 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
The value of the integral ∫ 0 π / 4 cos x 1 − sin 2 x d x \displaystyle\int_0^{\pi/4}\dfrac{\cos x}{\sqrt{1 - \sin^2 x}}\,dx ∫ 0 π /4 1 − sin 2 x cos x d x is #9 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If ∫ 0 x f ( t ) d t = x + ∫ x 0 t ⋅ f ( t ) d t \int_0^x f(t)\,dt = x + \int_x^0 t \cdot f(t)\,dt ∫ 0 x f ( t ) d t = x + ∫ x 0 t ⋅ f ( t ) d t then the value of f ( 1 ) = f(1) = f ( 1 ) = #10 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If a > 4 a > 4 a > 4 and ∫ − 2 1 1 x 2 + 4 x + a d x = 1 3 tan − 1 ( 1 ) \int_{-2}^1\dfrac{1}{x^2 + 4x + a}\,dx = \dfrac{1}{3}\tan^{-1}(1) ∫ − 2 1 x 2 + 4 x + a 1 d x = 3 1 tan − 1 ( 1 ) , then the value of a a a is #11 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If ∫ 0 ∞ 1 ( x 2 + 1 ) ( x 2 + 16 ) d x = π k 2 − 9 \int_0^\infty\dfrac{1}{(x^2 + 1)(x^2 + 16)}\,dx = \dfrac{\pi}{k^2 - 9} ∫ 0 ∞ ( x 2 + 1 ) ( x 2 + 16 ) 1 d x = k 2 − 9 π , then k = … k = \ldots k = … #12 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If ∫ − 1 4 f ( x ) d x = 4 \int_{-1}^4 f(x)dx = 4 ∫ − 1 4 f ( x ) d x = 4 and ∫ 2 4 ( 5 − f ( x ) ) d x = 7 \int_2^4(5 - f(x))dx = 7 ∫ 2 4 ( 5 − f ( x )) d x = 7 then ∫ 2 − 1 f ( x ) d x = \int_2^{-1}f(x)dx = ∫ 2 − 1 f ( x ) d x = #13 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
The values of a a a for which ∫ 0 a ( 3 x 2 + 4 x − 5 ) d x < a 3 − 2 \int_0^a(3x^2 + 4x - 5)\,dx < a^3 - 2 ∫ 0 a ( 3 x 2 + 4 x − 5 ) d x < a 3 − 2 are... A 1 2 < a < 2 \dfrac{1}{2} < a < 2 2 1 < a < 2 B 1 2 ≤ a ≤ 2 \dfrac{1}{2} \le a \le 2 2 1 ≤ a ≤ 2 C a ≤ 1 2 a \le \dfrac{1}{2} a ≤ 2 1 D #14 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
The value of the integral ∫ 10 20 1 ( x − 3 ) ( x − 2 ) d x \int_{10}^{20}\dfrac{1}{(x - 3)(x - 2)}\,dx ∫ 10 20 ( x − 3 ) ( x − 2 ) 1 d x is... A log ( 34 21 ) \log\left(\dfrac{34}{21}\right) log ( 21 34 ) B log ( 17 7 ) \log\left(\dfrac{17}{7}\right) log ( 7 17 ) C log ( 17 18 ) \log\left(\dfrac{17}{18}\right) log ( 18 17 ) D log ( 68 63 ) \log\left(\dfrac{68}{63}\right) log ( 63 68 ) #15 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
The value of the integral ∫ 8 15 d x ( x − 3 ) x + 1 \int_8^{15}\dfrac{dx}{(x - 3)\sqrt{x + 1}} ∫ 8 15 ( x − 3 ) x + 1 d x is.... A 1 2 log 5 3 \dfrac{1}{2}\log\dfrac{5}{3} 2 1 log 3 5 B 1 3 log 5 3 \dfrac{1}{3}\log\dfrac{5}{3} 3 1 log 3 5 C 1 2 log 3 5 \dfrac{1}{2}\log\dfrac{3}{5} 2 1 log 5 3 D 1 5 log 3 5 \dfrac{1}{5}\log\dfrac{3}{5} 5 1 log 5 3 MHT-CET · 2026 · 12 May Shift I · Q81 Moderate Add #16 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
∫ 0 4 4 − x 4 + x d x = \int_0^4\sqrt{\dfrac{4 - x}{4 + x}}\,dx = ∫ 0 4 4 + x 4 − x d x = MHT-CET · 2026 · 12 May Shift I · Q82 Moderate Add #17 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If n n n is an odd natural number and I n = ∫ 0 1 e x ( x − 1 ) n d x I_n = \int_0^1 e^x(x - 1)^n\,dx I n = ∫ 0 1 e x ( x − 1 ) n d x , then the value of I n + n I n − 1 I_n + nI_{n-1} I n + n I n − 1 is equal to ... #18 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If ∫ 0 1 d x x + 1 − x = 2 k \int_0^1\dfrac{dx}{\sqrt{x + 1} - \sqrt{x}} = \sqrt{2}k ∫ 0 1 x + 1 − x d x = 2 k , then the value of k k k is... A 4 3 \dfrac{4}{\sqrt{3}} 3 4 B C 2 3 \dfrac{2}{\sqrt{3}} 3 2 D #19 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
The value of integral ∫ 0 ∞ 1 1 + e x d x \int_0^\infty\dfrac{1}{1 + e^x}\,dx ∫ 0 ∞ 1 + e x 1 d x is... A B log ( 2 e ) \log\left(\dfrac{2}{e}\right) log ( e 2 ) C D log ( 4 e ) \log\left(\dfrac{4}{e}\right) log ( e 4 ) #20 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If f ( x ) f(x) f ( x ) is an antiderivative of x + 1 x − 1 \dfrac{x + 1}{\sqrt{x - 1}} x − 1 x + 1 with respect to x x x , then value of f ( 5 ) − f ( 2 ) f(5) - f(2) f ( 5 ) − f ( 2 ) is... #21 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If ∫ 0 2 x ( 2 − x ) b d x = 32 7 \int_0^2 x(2 - x)^b\,dx = \dfrac{32}{7} ∫ 0 2 x ( 2 − x ) b d x = 7 32 , where b ∈ N b \in N b ∈ N then b = b = b = MHT-CET · 2026 · 18 April Shift I · Q82 Hard Add #22 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
The value of the definite integral ∫ 0 log e 5 e x e x − 1 e x + 3 d x \int_0^{\log_e 5}\dfrac{e^x\sqrt{e^x - 1}}{e^x + 3}\,dx ∫ 0 l o g e 5 e x + 3 e x e x − 1 d x is equal to... #23 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If I n = ∫ 1 e ( log e x ) n d x I_n = \int_1^e(\log_e x)^n\,dx I n = ∫ 1 e ( log e x ) n d x where n ∈ Z + n \in Z^+ n ∈ Z + and I m + m I 2026 = e I_m + mI_{2026} = e I m + m I 2026 = e , then m = m = m = #24 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
∫ 0 log 5 e x e x − 1 e x + 3 d x = \displaystyle\int_0^{\log 5}\dfrac{e^x\sqrt{e^x - 1}}{e^x + 3}\,dx = ∫ 0 l o g 5 e x + 3 e x e x − 1 d x = A 2 − π 4 2 - \dfrac{\pi}{4} 2 − 4 π B 4 − π 4 4 - \dfrac{\pi}{4} 4 − 4 π C D #25 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If ∫ 0 a ( x 2 − 4 x + 1 ) d x = 6 \int_0^a(x^2 - 4x + 1)\,dx = 6 ∫ 0 a ( x 2 − 4 x + 1 ) d x = 6 , then the real value of a a a is #26 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If ∫ 0 a d x 1 + 4 x 2 = π 8 \displaystyle\int_0^a\dfrac{dx}{1 + 4x^2} = \dfrac{\pi}{8} ∫ 0 a 1 + 4 x 2 d x = 8 π , then a = a = a = #27 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If ∫ 2 e [ 1 log x − 1 ( log x ) 2 ] d x = a + b log 2 \displaystyle\int_2^e\left[\dfrac{1}{\log x} - \dfrac{1}{(\log x)^2}\right]dx = a + \dfrac{b}{\log 2} ∫ 2 e [ log x 1 − ( log x ) 2 1 ] d x = a + log 2 b , then values of a a a and b b b are A a = e , b = 2 a = e, b = 2 a = e , b = 2 B a = − e , b = 2 a = -e, b = 2 a = − e , b = 2 C a = e , b = − 2 a = e, b = -2 a = e , b = − 2 D a = − e , b = − 2 a = -e, b = -2 a = − e , b = − 2 #28 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If ∫ 0 k 1 1 + 4 x d x = log 2 ( 4 3 ) \displaystyle\int_0^k\dfrac{1}{1 + 4^x}\,dx = \log_2\left(\dfrac{4}{3}\right) ∫ 0 k 1 + 4 x 1 d x = log 2 ( 3 4 ) , then k = k = k = _____ A 2 3 \dfrac{2}{\sqrt{3}} 3 2 B C 3 2 \dfrac{\sqrt{3}}{2} 2 3 D #29 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If ∫ 1 a ( 2 x + 1 ) d x = 5 \displaystyle\int_1^a(2x + 1)\,dx = 5 ∫ 1 a ( 2 x + 1 ) d x = 5 , then the sum of the values of a a a is.. #30 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
∫ 0 log 10 [ e x e x − 1 e x + 8 ] d x = \displaystyle\int_0^{\log 10}\left[\dfrac{e^x\sqrt{e^x - 1}}{e^x + 8}\right]dx = ∫ 0 l o g 10 [ e x + 8 e x e x − 1 ] d x = A 3 2 ( 4 + π ) \dfrac{3}{2}(4 + \pi) 2 3 ( 4 + π ) B 3 2 ( 4 − π ) \dfrac{3}{2}(4 - \pi) 2 3 ( 4 − π ) C 3 4 ( 4 − π ) \dfrac{3}{4}(4 - \pi) 4 3 ( 4 − π ) D 3 4 ( 4 + π ) \dfrac{3}{4}(4 + \pi) 4 3 ( 4 + π ) #31 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
∫ 0 1 tan − 1 x d x = \int_{0}^{1} \tan^{- 1}x\text{ }dx= ∫ 0 1 tan − 1 x d x = A π 4 − log 2 \frac{\pi}{4} - \log2 4 π − log 2 B π 4 − log 2 \frac{\pi}{4} - \log\sqrt{2} 4 π − log 2 C π 4 + log 2 \frac{\pi}{4} + \log2 4 π + log 2 D π 4 + log 2 \frac{\pi}{4} + \log\sqrt{2} 4 π + log 2 MHT-CET · 2025 · 25 April Shift I · Q147 Hard Add #32 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
∫ 0 1 1 2 + x d x = \int_{0}^{1} \frac{1}{2 +\sqrt{x}}\text{ }dx= ∫ 0 1 2 + x 1 d x = A 2 log ( 2 e 3 ) \ 2\log\left( \frac{2e}{3} \right) 2 log ( 3 2 e ) B 2 log ( 4 e 9 ) \ 2\log\left( \frac{4e}{9} \right) 2 log ( 9 4 e ) C log ( 2 e 3 ) \ \log\left( \frac{2e}{3} \right) log ( 3 2 e ) D log ( 4 e 9 ) \ \log\left( \frac{4e}{9} \right) log ( 9 4 e ) #33 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
The value of the integral ∫ 1 2 x d x ( x + 2 ) ( x + 3 ) \int_{1}^{2} \frac{xdx}{(x + 2)(x + 3)} ∫ 1 2 ( x + 2 ) ( x + 3 ) x d x is A log ( 125 16 ) \ \log\left( \frac{125}{16} \right) log ( 16 125 ) B log ( 1024 1125 ) \ \log\left( \frac{1024}{1125} \right) log ( 1125 1024 ) C log ( 16 125 ) \ \log\left( \frac{16}{125} \right) log ( 125 16 ) D log ( 1125 1024 ) \log\left( \frac{1125}{1024} \right) log ( 1024 1125 ) #34 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
∫ 0 2 3 x + 1 x 2 + 4 d x = \int_{0}^{2} \frac{3x+ 1}{x^{2}+ 4}dx= ∫ 0 2 x 2 + 4 3 x + 1 d x = A log ( 2 2 ) + π 4 \ \log(2\sqrt{2}) + \frac{\pi}{4} log ( 2 2 ) + 4 π B log ( 2 2 ) + π 6 \ \log(2\sqrt{2}) + \frac{\pi}{6} log ( 2 2 ) + 6 π C log ( 2 2 ) + π 8 \ \log(2\sqrt{2}) + \frac{\pi}{8} log ( 2 2 ) + 8 π D log ( 2 2 ) + π 12 \ \log(2\sqrt{2}) + \frac{\pi}{12} log ( 2 2 ) + 12 π MHT-CET · 2025 · 20 April Shift II · Q121 Moderate Add #35 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
∫ 0 3 d x ( x + 2 ) x + 1 = \int_{0}^{3} \frac{dx}{(x+ 2)\sqrt{x+ 1}}= ∫ 0 3 ( x + 2 ) x + 1 d x = A tan − 1 ( 1 3 ) \tan^{- 1}\left( \frac{1}{3} \right) tan − 1 ( 3 1 ) B 2 tan − 1 ( 1 3 ) 2\tan^{- 1}\left( \frac{1}{3} \right) 2 tan − 1 ( 3 1 ) C 3 tan − 1 ( 1 3 ) 3\tan^{- 1}\left( \frac{1}{3} \right) 3 tan − 1 ( 3 1 ) D 4 tan − 1 ( 1 3 ) 4\tan^{- 1}\left( \frac{1}{3} \right) 4 tan − 1 ( 3 1 ) #36 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
The value of ∫ 1 3 1 ( x − x 3 ) 1 3 x 4 d x \int_{\frac{1}{3}}^{1} \frac{\left( x-x^{3} \right)^{\frac{1}{3}}}{x^{4}}dx ∫ 3 1 1 x 4 ( x − x 3 ) 3 1 d x is MHT-CET · 2025 · 19 April Shift I · Q124 Moderate Add #37 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
∫ 1 e e x x ( 1 + x l o g x ) d x = \int_{1}^{e} \frac{e^{x}}{x}(1 +xlogx)dx= ∫ 1 e x e x ( 1 + x l o g x ) d x = MHT-CET · 2023 · 10th May Shift 2 · Q121 Moderate Add #38 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
∫ 0 a x − a x + a d x = \displaystyle\int_0^a \frac{x-a}{x+a}\,dx = ∫ 0 a x + a x − a d x = A a − 2 a log 2 a - 2a\log 2 a − 2 a log 2 B a + 2 a log 2 a + 2a\log 2 a + 2 a log 2 C a − a log 2 a - a\log 2 a − a log 2 D a + a log 2 a + a\log 2 a + a log 2 #39 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If ∫ 0 π / 3 tan θ 2 k sec θ d θ = 1 − 1 2 , ( k > 0 ) \int_0^{\pi/3}\frac{\tan\theta}{\sqrt{2k\sec\theta}}\,d\theta=1-\frac{1}{\sqrt{2}},\ (k>0) ∫ 0 π /3 2 k s e c θ t a n θ d θ = 1 − 2 1 , ( k > 0 ) , then the value of k k k is #40 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If I n = ∫ 0 π / 4 tan n θ d θ I_n = \int_0^{\pi/4} \tan^n\theta\,d\theta I n = ∫ 0 π /4 tan n θ d θ , then I 12 + I 10 I_{12} + I_{10} I 12 + I 10 = #41 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
∫ 0 1 cos − 1 x d x = \displaystyle\int_0^1 \cos^{-1}x\,dx= ∫ 0 1 cos − 1 x d x = #42 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If ∫ 0 1 / 2 x 2 ( 1 − x 2 ) 3 / 2 d x = k 6 \int_0^{1/2}\frac{x^2}{(1-x^2)^{3/2}}dx=\frac{k}{6} ∫ 0 1/2 ( 1 − x 2 ) 3/2 x 2 d x = 6 k , value of k is #43 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
∫ 1 2 d x ( x 2 − 2 x + 4 ) 3 / 2 = k k + 5 \int_1^2 \dfrac{dx}{\left(x^2 - 2x + 4\right)^{3/2}} = \dfrac{k}{k+5} ∫ 1 2 ( x 2 − 2 x + 4 ) 3/2 d x = k + 5 k , then k k k has the value MHT-CET · 2024 · 9th May Shift 1 · Q134 Hard Add #44 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
If f ( x ) f(x) f ( x ) is a function satisfying f ′ ( x ) = f ( x ) f'(x)=f(x) f ′ ( x ) = f ( x ) with f ( 0 ) = 1 f(0)=1 f ( 0 ) = 1 and g ( x ) g(x) g ( x ) is a function that satisfies f ( x ) + g ( x ) = x 2 f(x)+g(x)=x^2 f ( x ) + g ( x ) = x 2 . Then the value of the integral ∫ 0 1 f ( x ) g ( x ) d x \int_0^1 f(x)g(x)\,dx ∫ 0 1 f ( x ) g ( x ) d x is A e − e 2 2 − 5 2 e-\frac{e^2}{2}-\frac{5}{2} e − 2 e 2 − 2 5 B e + e 2 2 − 3 2 e+\frac{e^2}{2}-\frac{3}{2} e + 2 e 2 − 2 3 C e − e 2 2 − 3 2 e-\frac{e^2}{2}-\frac{3}{2} e − 2 e 2 − 2 3 D e + e 2 2 + 5 2 e+\frac{e^2}{2}+\frac{5}{2} e + 2 e 2 + 2 5 MHT-CET · 2022 · Shift 1 · Q127 Hard Add #45 · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts
The value of the integral ∫ 0 1 1 − x 1 + x d x \int_0^1 \sqrt{\frac{1-x}{1+x}}\,dx ∫ 0 1 1 + x 1 − x d x is A π 2 + 1 \frac{\pi}{2} + 1 2 π + 1 B π 2 − 1 \frac{\pi}{2} - 1 2 π − 1 C D #46 · Trigonometric Definite Integrals — tan x = t, Half-Angle Forms and Powers
The value of ∫ 0 π / 2 cos 3 x + 1 2 cos x − 1 d x \displaystyle\int_0^{\pi/2} \dfrac{\cos 3x + 1}{2\cos x - 1}dx ∫ 0 π /2 2 cos x − 1 cos 3 x + 1 d x is #47 · Trigonometric Definite Integrals — tan x = t, Half-Angle Forms and Powers
∫ 0 π 2 ( 1 + sin x ) 3 d x = \int_0^{\frac{\pi}{2}}(1 + \sin x)^3\,dx = ∫ 0 2 π ( 1 + sin x ) 3 d x = A B C π 2 − 2 3 \dfrac{\pi}{2} - \dfrac{2}{3} 2 π − 3 2 D 5 π 4 + 11 3 \dfrac{5\pi}{4} + \dfrac{11}{3} 4 5 π + 3 11 #48 · Trigonometric Definite Integrals — tan x = t, Half-Angle Forms and Powers
If f ( x ) = ∣ sin x + sin 2 x + sin 3 x sin 2 x sin 3 x 3 + 4 sin x 3 4 sin x 1 + sin x sin x 1 ∣ f(x) = \begin{vmatrix} \sin x + \sin 2x + \sin 3x & \sin 2x & \sin 3x \\ 3 + 4\sin x & 3 & 4\sin x \\ 1 + \sin x & \sin x & 1 \end{vmatrix} f ( x ) = sin x + sin 2 x + sin 3 x 3 + 4 sin x 1 + sin x sin 2 x 3 sin x sin 3 x 4 sin x 1 , then ∫ 0 π 2 f ( x ) d x = \int_0^{\frac{\pi}{2}}f(x)\,dx = ∫ 0 2 π f ( x ) d x = #49 · Trigonometric Definite Integrals — tan x = t, Half-Angle Forms and Powers
If I n = ∫ 0 π / 4 tan n x d x , n ∈ N I_n = \int_0^{\pi/4}\tan^n x\,dx, n \in N I n = ∫ 0 π /4 tan n x d x , n ∈ N then I n + 2 + I n I_{n+2} + I_n I n + 2 + I n is equal to A B 1 n + 1 \dfrac{1}{n + 1} n + 1 1 C 1 n − 1 \dfrac{1}{n - 1} n − 1 1 D 1 n − 2 \dfrac{1}{n - 2} n − 2 1 #50 · Trigonometric Definite Integrals — tan x = t, Half-Angle Forms and Powers
The value of the definite integral ∫ 0 π 1 5 + 4 cos x d x \int_0^\pi\dfrac{1}{5 + 4\cos x}\,dx ∫ 0 π 5 + 4 cos x 1 d x is equal to ... Showing the first 50 of 132.
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