Integral Calculus jee mains questions | JEE Main Maths Integral Calculus Previous Year Question

Q 1 :
$If \int \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x = \frac{1}{12} \tan^{- 1} (3 \tan x) + constant,$ $then the maximum value of a \sin x + b \cos x is:$ [2024]

$\sqrt{41}$
$\sqrt{42}$
$\sqrt{40}$
$\sqrt{39}$

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(3)

We have, $\int \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x = \frac{1}{12} \tan^{- 1} (3 \tan x) + c$

Let $I = \int \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x = \int \frac{\sec^{2} x}{a^{2} \tan^{2} x + b^{2}} d x$

Let $\tan x = t \Rightarrow \sec^{2} x d x = d t$

$∴$ $I = \int \frac{d t}{{(a t)}^{2} + b^{2}}$

$= \frac{1}{b a} \tan^{- 1} (\frac{a t}{b}) + c = \frac{1}{b a} \tan^{- 1} (\frac{a}{b} \tan x) + c$

$\Rightarrow a b = 12 and \frac{a}{b} = 3$

$\Rightarrow a^{2} = 36 and b^{2} = 4$

Now, $- \sqrt{a^{2} + b^{2}} \leq a \sin x + b \cos x \leq \sqrt{a^{2} + b^{2}}$

Required maximum value $= \sqrt{36 + 4} = \sqrt{40}$

Q 2 :
Let $I (x) = \int \frac{6}{\sin^{2} x {(1 - \cot x)}^{2}} d x .$ If $I (0) = 3,$ then $I (\frac{π}{12})$ is equal to [2024]

$6 \sqrt{3}$
$3 \sqrt{3}$
$\sqrt{3}$
$2 \sqrt{3}$

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(2)

$I (x) = \int \frac{6}{\sin^{2} x {(1 - \cot x)}^{2}} d x = \int \frac{6 c o s e c^{2} x}{{(1 - \cot x)}^{2}} d x$

Put $1 - \cot x = t \Rightarrow c o s e c^{2} x d x = d t$

$\Rightarrow I = \int \frac{6 d t}{t^{2}} = \frac{- 6}{t} + C \Rightarrow I = \frac{- 6}{1 - \cot x} + C$

Now, $I (0) = 3 \Rightarrow 3 = C$

$\Rightarrow I (x) = 3 - \frac{6}{1 - \cot x} \Rightarrow I (\frac{π}{12}) = 3 - \frac{6}{1 - \cot \frac{π}{12}}$

$= 3 - \frac{6}{1 - (2 + \sqrt{3})} = \frac{- 3 - 3 \sqrt{3} - 6}{- 1 - \sqrt{3}} = \frac{- 9 - 3 \sqrt{3}}{- 1 - \sqrt{3}}$

$= \frac{(9 + 3 \sqrt{3}) (1 - \sqrt{3})}{- 2} = \frac{9 - 9 \sqrt{3} + 3 \sqrt{3} - 9}{- 2} = 3 \sqrt{3}$

Q 3 :
Let $\int \frac{2 - \tan x}{3 + \tan x} d x = \frac{1}{2} (α x + \log_{e} | β \sin x + γ \cos x |) + C,$ where $C$ is the constant of integration. Then $α + \frac{γ}{β}$ is equal to: [2024]

7
3
1
4

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(4)

Let $I = \int \frac{2 - \tan x}{3 + \tan x} d x = \int \frac{2 \cos x - \sin x}{3 \cos x + \sin x} d x$

$Write Numerator = λ (Differentiation of Denominator) + μ (Denominator)$

$2 \cos x - \sin x = λ (- 3 \sin x + \cos x) + μ (3 \cos x + \sin x)$

$\Rightarrow λ + 3 μ = 2 \dots (i)$

$and - 3 λ + μ = - 1 \dots (ii)$

$From equations (i) and (ii), λ = μ = \frac{1}{2}$

$I = \frac{1}{2} \int \frac{- 3 \sin x + \cos x}{3 \cos x + \sin x} d x + \frac{1}{2} \int \frac{3 \cos x + \sin x}{3 \cos x + \sin x} d x$

$= \frac{1}{2} \int \frac{d t}{t} + \frac{1}{2} \int d x [Putting 3 \cos x + \sin x = t in first integral \Rightarrow (- 3 \sin x + \cos x) d x = d t]$

$= \frac{1}{2} \ln (t) + \frac{1}{2} x + C$

$= \frac{1}{2} \ln$ $| 3 \cos x + \sin x |$ $+ \frac{1}{2} x + C$

$= \frac{1}{2} (α x + \log_{e} | β \sin x + γ \cos x |) + C (∵Given)$

$\Rightarrow α = 1, β = 1, γ = 3$

$Now, α + \frac{γ}{β} = 1 + \frac{3}{1} = 4$

Q 4 :
The integral $\int \frac{(x^{8} - x^{2}) d x}{(x^{12} + 3 x^{6} + 1) \tan^{- 1} (x^{3} + \frac{1}{x^{3}})}$ is equal to: [2024]

$\log_{e}$ $(| \tan^{- 1} (x^{3} + \frac{1}{x^{3}}) |) + C$
$\log_{e}$ $(| \tan^{- 1} {(x^{3} + \frac{1}{x^{3}}) |)}^{3} + C$
$\log_{e}$ $(| \tan^{- 1} {(x^{3} + \frac{1}{x^{3}}) |)}^{\frac{1}{3}} + C$
$\log_{e}$ $(| \tan^{- 1} {(x^{3} + \frac{1}{x^{3}}) |)}^{\frac{1}{2}} + C$

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(3)

Let $I = \int \frac{(x^{8} - x^{2}) d x}{(x^{12} + 3 x^{6} + 1) \tan^{- 1} (x^{3} + \frac{1}{x^{3}})}$

Put $\tan^{- 1} (x^{3} + \frac{1}{x^{3}}) = u$

$\Rightarrow \frac{1}{1 + {(x^{3} + \frac{1}{x^{3}})}^{2}} \cdot 3 (x^{2} - \frac{1}{x^{4}}) d x = d u$

$\Rightarrow (\frac{x^{6}}{x^{12} + 3 x^{6} + 1} \times \frac{x^{6} - 1}{x^{4}}) d x = \frac{d u}{3}$

$\Rightarrow I = \frac{1}{3} \int \frac{d u}{u} = \frac{\ln | u |}{3} = \ln$ $(| \tan^{- 1} {(x^{3} + \frac{1}{x^{3}}) |)}^{\frac{1}{3}} + C$

Q 5 :
$For x \in (- \frac{π}{2}, \frac{π}{2}), if y (x) = \int \frac{c o s e c x + \sin x}{c o s e c x \sec x + \tan x \sin^{2} x} d x,$ $and \lim_{x \to {(\frac{π}{2})}^{-}} y (x) = 0, then y (\frac{π}{4}) is equal to$ [2024]

$- \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{1}{\sqrt{2}})$
$\tan^{- 1} (\frac{1}{\sqrt{2}})$
$\frac{1}{2} \tan^{- 1} (\frac{1}{\sqrt{2}})$
$\frac{1}{\sqrt{2}} \tan^{- 1} (- \frac{1}{2})$

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(4)

Let $I = y (x) = \int \frac{c o s e c x + \sin x}{c o s e c x \sec x + \tan x \sin^{2} x} d x$

$= \int \frac{\frac{1}{\sin x} + \sin x}{\frac{1}{\sin x} \cdot \frac{1}{\cos x} + \frac{\sin x}{\cos x} \cdot \sin^{2} x} d x = \int \frac{(1 + \sin^{2} x) \cos x}{(1 + \sin^{4} x)} d x$

Let $\sin x = t \Rightarrow \cos x d x = d t$

$∴ I = \int \frac{1 + t^{2}}{1 + t^{4}} d t = \int \frac{(1 + \frac{1}{t^{2}})}{(t^{2} + \frac{1}{t^{2}})} d t$

Put $t - \frac{1}{t} = u \Rightarrow (1 + \frac{1}{t^{2}}) d t = d u and t^{2} + \frac{1}{t^{2}} = u^{2} + 2$

$∴ I = \int \frac{d u}{u^{2} + 2} = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{u}{\sqrt{2}}) + C = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{1}{\sqrt{2}} (t - \frac{1}{t}) + C$

$y (x) = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{1}{\sqrt{2}} (\sin x - \cos e c x) + C$

$\lim_{x \to {(\frac{π}{2})}^{-}} y (x) = \lim_{h \to 0} [\frac{1}{\sqrt{2}} \tan^{- 1} \frac{1}{\sqrt{2}} {\sin (\frac{π}{2} - h) - c o s e c (\frac{π}{2} - h)} + c]$

$\Rightarrow \frac{1}{\sqrt{2}} \tan^{- 1} \frac{1}{\sqrt{2}} (0) + c = 0 \Rightarrow c = 0$

$∴ y (\frac{π}{4}) = \frac{1}{\sqrt{2}} \tan^{- 1} {\frac{1}{\sqrt{2}} (\sin \frac{π}{4} - c o s e c \frac{π}{4})}$

$= \frac{1}{\sqrt{2}} \tan^{- 1} {\frac{1}{\sqrt{2}} (\frac{1}{\sqrt{2}} - \sqrt{2})} = \frac{1}{\sqrt{2}} \tan^{- 1} (- \frac{1}{2})$

Q 6 :
If $\int \frac{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x}{\sqrt{\sin^{3} x \cos^{3} x \sin (x - θ)}} d x = A \sqrt{\cos θ \tan x - \sin θ} + B \sqrt{\cos θ - \sin θ \cot x} + C,$ where C is the integration constant, then AB is equal to [2024]

$4 c o s e c (2 θ)$
$2 \sec θ$
$4 \sec θ$
$8 c o s e c (2 θ)$

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(4)

Let $I = \int \frac{\sin^{3 / 2} x + \cos^{3 / 2} x}{\sqrt{\sin^{3} x \cdot \cos^{3} x \cdot \sin (x - θ)}} d x$

$= \int \frac{d x}{\sqrt{\cos^{3} x \cdot \sin (x - θ)}} + \int \frac{d x}{\sqrt{\sin^{3} x \cdot \sin (x - θ)}}$

$= \int \frac{d x}{\cos^{2} x \sqrt{\frac{\sin x \cos θ - \cos x \sin θ}{\cos x}}} + \int \frac{d x}{\sin^{2} x \sqrt{\frac{\sin x \cos θ - \cos x \sin θ}{\sin x}}}$

$= \int \frac{\sec^{2} x}{\sqrt{\tan x \cos θ - \sin θ}} d x + \int \frac{c o s e c^{2} x}{\sqrt{\cos θ - \cot x \sin θ}} d x$

$\Rightarrow I = I_{1} + I_{2} (Let)$ ...(i)

Now, $I_{1} = \int \frac{\sec^{2} x d x}{\sqrt{\tan x \cos θ - \sin θ}}$

Put $\tan x \cos θ - \sin θ = t \Rightarrow \cos θ \cdot \sec^{2} x d x = d t$

$∴ I_{1} = \int \frac{d t}{\cos θ \cdot \sqrt{t}} \Rightarrow I_{1} = \frac{2 t^{1 / 2}}{\cos θ}$

$\Rightarrow I_{1} = \frac{2 {(\tan x \cos θ - \sin θ)}^{1 / 2}}{\cos θ}$

And, $I_{2} = \int \frac{c o s e c^{2} x}{\sqrt{\cos θ - \cot x \sin θ}} d x$

Put $\cos θ - \cot x \sin θ = v \Rightarrow \sin θ c o s e c^{2} x d x = d v$

$∴ I_{2} = \int \frac{d v}{\sin θ \sqrt{v}} = \frac{2 \sqrt{v}}{\sin θ} \Rightarrow I_{2} = \frac{2 \sqrt{\cos θ - \cot x \sin θ}}{\sin θ}$

From (i),

$I = \frac{2 \sqrt{\tan x \cos θ - \sin θ}}{\cos θ} + \frac{2 \sqrt{\cos θ - \cot x \sin θ}}{\sin θ} + C$ and

$I = A \sqrt{\cos θ \tan x - \sin θ} + B \sqrt{\cos θ - \sin θ \cot x} + C$

So, $A = \frac{2}{\cos θ}, B = \frac{2}{\sin θ}$

$∴ A B = \frac{4 \cdot 2}{2 \sin θ \cos θ} = 8 c o s e c 2 θ$

Q 7 :
If $\int c o s e c^{5} x d x = α \cot x c o s e c x (c o s e c^{2} x + \frac{3}{2}) + β \log_{e} | \tan \frac{x}{2} | + C,$ where $α, β \in R$ and $C$ is the constant of integration, then the value of $8 (α + β)$ equals _______ . [2024]

0%

(1)

$I_{n} = \int c o s e c^{n} x d x$

Using reduction formula, we get

$I_{n} = \frac{- c o s e c^{n - 2} x \cot x}{n - 1} + \frac{n - 2}{n - 1} I_{n - 2}$

$∴$ $I_{5} = \int c o s e c^{5} x d x$

$= \frac{- c o s e c^{3} x \cot x}{4} + \frac{3}{4} I_{3} + C$

$= \frac{- c o s e c^{3} x \cot x}{4} + \frac{3}{4} [\frac{- c o s e c x \cot x}{2} + \frac{1}{2} \int c o s e c x d x] + C$

$= \frac{- 1}{4} c o s e c^{3} x \cot x - \frac{3}{8} c o s e c x \cot x + \frac{3}{8} \log_{e} | \tan \frac{x}{2} | + C$

$= \frac{- 1}{4} c o s e c x \cot x [c o s e c^{2} x + \frac{3}{2}] + \frac{3}{8} \log_{e} | \tan \frac{x}{2} | + C$

$= α \cot x c o s e c x [c o s e c^{2} x + \frac{3}{2}] + β \log_{e} | \tan \frac{x}{2} | + C (∵ Given)$

On comparing, we get $α = \frac{- 1}{4}$ and $β = \frac{3}{8}$

Hence, $8 (α + β) = 8 (\frac{- 1}{4} + \frac{3}{8}) = 8 \times \frac{1}{8} = 1$

Q 8 :
$If \int \frac{1}{\sqrt[5]{{(x - 1)}^{4} {(x + 3)}^{6}}} d x = A {(\frac{α x - 1}{β x + 3})}^{B} + C, where C is the constant of integration,$ $then the value of α + β + 20 A B is ____ .$ [2024]

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(7)

$I = \int \frac{1}{\sqrt[5]{{(x - 1)}^{4} {(x + 3)}^{6}}} d x$

Putting $\frac{x + 3}{x - 1} = t \Rightarrow x = (\frac{3 + t}{t - 1}) \Rightarrow d x = \frac{- 4}{{(t - 1)}^{2}} d t$

$\Rightarrow {(x - 1)}^{4} {(x + 3)}^{6} = {(x - 1)}^{5} {(x + 3)}^{5} (\frac{x + 3}{x - 1})$

$∴ I = \int \frac{\frac{- 4}{{(t - 1)}^{2}} d t}{t^{1 / 5} (\frac{3 + t}{t - 1} - 1) (\frac{3 + t}{t - 1} + 3)}$

$I = \int \frac{- 4 d t}{t^{1 / 5} (16 t)} = \frac{- 1}{4} \int \frac{d t}{t^{6 / 5}} = \frac{- 1}{4} (\frac{t^{- 1 / 5}}{- 1 / 5}) + C$

On comparing, we get

$A = \frac{5}{4}, B = \frac{1}{5}, α = 1, β = 1$

Hence, $α + β + 20 A B = 1 + 1 + 20 \times \frac{5}{4} \times \frac{1}{5} = 7$

Topic Question Set

Q 1 :
$If \int \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x = \frac{1}{12} \tan^{- 1} (3 \tan x) + constant,$ $then the maximum value of a \sin x + b \cos x is:$ [2024]

Q 2 :
Let $I (x) = \int \frac{6}{\sin^{2} x {(1 - \cot x)}^{2}} d x .$ If $I (0) = 3,$ then $I (\frac{π}{12})$ is equal to [2024]

Q 3 :
Let $\int \frac{2 - \tan x}{3 + \tan x} d x = \frac{1}{2} (α x + \log_{e} | β \sin x + γ \cos x |) + C,$ where $C$ is the constant of integration. Then $α + \frac{γ}{β}$ is equal to: [2024]

Q 4 :
The integral $\int \frac{(x^{8} - x^{2}) d x}{(x^{12} + 3 x^{6} + 1) \tan^{- 1} (x^{3} + \frac{1}{x^{3}})}$ is equal to: [2024]

Q 5 :
$For x \in (- \frac{π}{2}, \frac{π}{2}), if y (x) = \int \frac{c o s e c x + \sin x}{c o s e c x \sec x + \tan x \sin^{2} x} d x,$ $and \lim_{x \to {(\frac{π}{2})}^{-}} y (x) = 0, then y (\frac{π}{4}) is equal to$ [2024]

Q 6 :
If $\int \frac{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x}{\sqrt{\sin^{3} x \cos^{3} x \sin (x - θ)}} d x = A \sqrt{\cos θ \tan x - \sin θ} + B \sqrt{\cos θ - \sin θ \cot x} + C,$ where C is the integration constant, then AB is equal to [2024]

Q 7 :
If $\int c o s e c^{5} x d x = α \cot x c o s e c x (c o s e c^{2} x + \frac{3}{2}) + β \log_{e} | \tan \frac{x}{2} | + C,$ where $α, β \in R$ and $C$ is the constant of integration, then the value of $8 (α + β)$ equals _______ . [2024]

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Q 8 :
$If \int \frac{1}{\sqrt[5]{{(x - 1)}^{4} {(x + 3)}^{6}}} d x = A {(\frac{α x - 1}{β x + 3})}^{B} + C, where C is the constant of integration,$ $then the value of α + β + 20 A B is ____ .$ [2024]

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Topic Question Set

Q 1 : If ∫1a2sin2x+b2cos2x dx=112tan-1(3tanx)+constant, then the maximum value of asinx+bcosx is: [2024]

Q 2 : Let I(x)=∫6sin2x(1-cotx)2 dx. If I(0)=3, then I(π12) is equal to [2024]

Q 3 : Let ∫2-tanx3+tanx dx=12(αx+loge|βsinx+γcosx|)+C, where C is the constant of integration. Then α+γβ is equal to: [2024]

Q 4 : The integral ∫(x8-x2)dx(x12+3x6+1)tan-1(x3+1x3) is equal to: [2024]

Q 5 : For x∈(-π2,π2), if y(x)=∫cosecx+sinxcosecxsecx+tanxsin2x dx, and limx→(π2)-y(x)=0, then y(π4) is equal to [2024]

Q 6 : If ∫sin32x+cos32xsin3xcos3xsin(x-θ) dx=Acosθtanx-sinθ+Bcosθ-sinθcotx+C, where C is the integration constant, then AB is equal to [2024]

Q 7 : If ∫cosec5x dx=αcotxcosecx(cosec2x+32)+βloge|tanx2|+C, where α,β∈R and C is the constant of integration, then the value of 8(α+β) equals _______ . [2024]

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Q 8 : If ∫1(x-1)4(x+3)65 dx=A(αx-1βx+3)B+C, where C is the constant of integration, then the value of α+β+20AB is ____ . [2024]

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Q 1 :
$If \int \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x = \frac{1}{12} \tan^{- 1} (3 \tan x) + constant,$ $then the maximum value of a \sin x + b \cos x is:$ [2024]

Q 2 :
Let $I (x) = \int \frac{6}{\sin^{2} x {(1 - \cot x)}^{2}} d x .$ If $I (0) = 3,$ then $I (\frac{π}{12})$ is equal to [2024]

Q 3 :
Let $\int \frac{2 - \tan x}{3 + \tan x} d x = \frac{1}{2} (α x + \log_{e} | β \sin x + γ \cos x |) + C,$ where $C$ is the constant of integration. Then $α + \frac{γ}{β}$ is equal to: [2024]

Q 4 :
The integral $\int \frac{(x^{8} - x^{2}) d x}{(x^{12} + 3 x^{6} + 1) \tan^{- 1} (x^{3} + \frac{1}{x^{3}})}$ is equal to: [2024]

Q 5 :
$For x \in (- \frac{π}{2}, \frac{π}{2}), if y (x) = \int \frac{c o s e c x + \sin x}{c o s e c x \sec x + \tan x \sin^{2} x} d x,$ $and \lim_{x \to {(\frac{π}{2})}^{-}} y (x) = 0, then y (\frac{π}{4}) is equal to$ [2024]

Q 6 :
If $\int \frac{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x}{\sqrt{\sin^{3} x \cos^{3} x \sin (x - θ)}} d x = A \sqrt{\cos θ \tan x - \sin θ} + B \sqrt{\cos θ - \sin θ \cot x} + C,$ where C is the integration constant, then AB is equal to [2024]

Q 7 :
If $\int c o s e c^{5} x d x = α \cot x c o s e c x (c o s e c^{2} x + \frac{3}{2}) + β \log_{e} | \tan \frac{x}{2} | + C,$ where $α, β \in R$ and $C$ is the constant of integration, then the value of $8 (α + β)$ equals _______ . [2024]

Q 8 :
$If \int \frac{1}{\sqrt[5]{{(x - 1)}^{4} {(x + 3)}^{6}}} d x = A {(\frac{α x - 1}{β x + 3})}^{B} + C, where C is the constant of integration,$ $then the value of α + β + 20 A B is ____ .$ [2024]