Evaluation of Special Integral Forms

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JEE Main 2012

If $\int \frac{5 t a n x}{t a n x - 2} d x = x + a ln ∣ sin x - 2 cos x ∣ + k$ , then $a$ is equal to

- 1

(B)

- 2

(C)

1

(D)

2

JEE Main 2005

LEVELJEE Main

$\int {\frac{l o g x - 1}{1 + ( l o g x ) ^{2}}}^{2} d x$ is equal to

(A)

\frac{l o g x}{( l o g x ) ^{2} + 1} + C

(B)

\frac{x}{x ^{2} + 1} + C

(C)

\frac{x e ^{x}}{1 + x ^{2}} + C

(D)

\frac{x}{( l o g x ) ^{2} + 1} + C

JEE Main 2004

LEVELJEE Main

$\int \frac{d x}{c o s x - s i n x}$ is equal to

(A)

\frac{1}{2} lo g tan (\frac{x}{2} + \frac{3 π}{8}) + C

(B)

\frac{1}{2} lo g cot (\frac{x}{2}) + C

(C)

\frac{1}{2} lo g tan (\frac{x}{2} - \frac{3 π}{8}) + C

(D)

\frac{1}{2} lo g tan (\frac{x}{2} - \frac{π}{8}) + C

JEE Advanced 2008

LEVELJEE Main

Let $I = \int \frac{e ^{x}}{e ^{4 x} + e ^{2 x} + 1} d x, J = \int \frac{e ^{- x}}{e ^{- 4 x} + e ^{- 2 x} + 1} d x$ . Then, for an arbitrary constant $C$ , the value of $J - I$ equals

(A)

\frac{1}{2} lo g \frac{e ^{4 x} - e ^{2 x} + 1}{e ^{4 x} + e ^{2 x} + 1} + C

(B)

\frac{1}{2} lo g \frac{e ^{2 x} + e ^{x} + 1}{e ^{2 x} - e ^{x} + 1} + C

(C)

\frac{1}{2} lo g \frac{e ^{2 x} - e ^{x} + 1}{e ^{2 x} + e ^{x} + 1} + C

(D)

\frac{1}{2} lo g \frac{e ^{4 x} + e ^{2 x} + 1}{e ^{4 x} - e ^{2 x} + 1} + C

JEE Advanced 1983

LEVELBoard

Evaluate $\int \frac{( x - 1 ) e ^{x}}{( x + 1 ) ^{3}} d x$

JEE Advanced 1978

LEVELJEE Main

Evaluate $\int \frac{s i n x}{s i n x - c o s x} d x$

JEE Main 2017

LEVELBoard

Let $I_{n} = \int tan^{n} x d x, (n > 1)$ . If $I_{4} + I_{6} = a tan^{5} x + b x^{5} + C$ , where C is a constant of integration, then the ordered pair $(a, b)$ is equal to:

(A)

(- \frac{1}{5}, 1)

(B)

(\frac{1}{5}, 0)

(C)

(\frac{1}{5}, - 1)

(D)

(- \frac{1}{5}, 0)

JEE Main 2018 (15 April Evening)

LEVELJEE Main

If $\int \frac{2 x + 5}{7 - 6 x - x ^{2}} d x = A 7 - 6 x - x^{2} + B sin^{- 1} (\frac{x + 3}{4}) + C$ (Where C is a constant of integration), then the ordered pair (A, B) is equal to :-

JEE Main 2018 (16 April Shift 1)

LEVELJEE Main

If $\int \frac{t a n x}{1 + t a n x + t a n ^{2} x} d x = x - \frac{K}{A} tan^{- 1} (\frac{K t a n x + 1}{A}) + C$ , (C is a constant of integration), then the ordered pair (K, A) is equal to :

(A)

(- 2, 3)

(B)

(- 2, 1)

(C)

(2, 1)

(D)

(2, 3)

JEE Main 2019 (10 April Shift 1)

LEVELJEE Main

If $\int \frac{d x}{( x ^{2} - 2 x + 10 ) ^{2}} = A (tan^{- 1} (\frac{x - 1}{3}) + \frac{f ( x )}{x ^{2} - 2 x + 10}) + C$ , where $C$ is a constant of integration, then :

(A)

A = 1/27

and

f (x) = 9 (x - 1)

(B)

A = 1/81

and

f (x) = 3 (x - 1)

(C)

A = 1/54

and

f (x) = 9 (x - 1)^{2}

(D)

A = 1/54

and

f (x) = 3 (x - 1)

JEE Main 2020 (5 September Shift 2)

LEVELJEE Main

If $\int \frac{c o s θ}{5 + 7 s i n θ - 2 c o s ^{2} θ} d θ = A lo g_{e} ∣ B (θ) ∣ + C$ where $C$ is a constant of integration, then $\frac{B ( θ )}{A}$ can be:

(A)

\frac{2 s i n θ + 1}{5 ( s i n θ + 3 )}

(B)

\frac{2 s i n θ + 1}{s i n θ + 3}

(C)

\frac{5 ( 2 s i n θ + 1 )}{s i n θ + 3}

(D)

\frac{5 ( s i n θ + 3 )}{2 s i n θ + 1}

JEE Main 2021 (27 Aug Shift 1)

LEVELJEE Main

If $\int \frac{d x}{( x ^{2} + x + 1 ) ^{2}} = a tan^{- 1} (\frac{2 x + 1}{3}) + b (\frac{2 x + 1}{x ^{2} + x + 1}) + C, x > 0$ where $C$ is the constant of integration, then the value of $9 (3 a + b)$ is equal to.

JEE Main 2021 (16 March Shift 2)

LEVELJEE Advanced

For real numbers $α, β, γ$ and $δ$ , if $\int \frac{(x^2 - 1) + \tan^{-1}\left(\frac{x^2+1}{x}\right)}{(x^4 + 3x^2 + 1)\tan^{-1}\left(\frac{x^2+1}{x}\right)} dx = \alpha \log_e\left(\tan^{-1}\left(\frac{x^2+1}{x}\right) ight) + \beta \tan^{-1}\left(\frac{\gamma(x^2-1)}{x}\right) + \delta \tan^{-1}\left(\frac{x^2+1}{x}\right) + C $w h er e$ C $i s ana r bi t r a r y co n s t an t, t h e n t h e v a l u eo f$ 10(\alpha + \beta\gamma + \delta)$ is equal to ____

JEE Main 2022 (27 June Shift 1)

LEVELJEE Main

$\int \frac{( x ^{2} + 1 ) e ^{x}}{( x + 1 ) ^{2}} d x = f (x) e^{x} + C$ , Where $C$ is constant, then $\frac{d ^{3} f}{d x ^{3}}$ at $x = 1$ is equal to :

(A)

- \frac{3}{4}

(B)

\frac{3}{4}

(C)

- \frac{3}{2}

(D)

\frac{3}{2}

JEE Main 2022 (26 July Shift 2)

LEVELJEE Main

The integral $\int \frac{( 1 - \frac{1}{3} ) ( c o s x - s i n x )}{1 + \frac{2}{3} s i n 2 x} d x$ is equal to

(A)

\frac{1}{2} lo g_{e} \frac{t a n ( \frac{x}{2} + \frac{π}{12} )}{t a n ( \frac{x}{2} + \frac{π}{6} )} + C

(B)

\frac{1}{2} lo g_{e} \frac{t a n ( \frac{x}{2} + \frac{π}{6} )}{t a n ( \frac{x}{2} + \frac{π}{3} )} + C

(C)

lo g_{e} \frac{t a n ( \frac{x}{2} + \frac{π}{6} )}{t a n ( \frac{x}{2} + \frac{π}{12} )} + C

(D)

\frac{1}{2} lo g_{e} \frac{t a n ( \frac{x}{2} - \frac{π}{12} )}{t a n ( \frac{x}{2} - \frac{π}{6} )} + C

JEE Main 2023 (08 April Shift 2)

LEVELJEE Advanced

The integral $\int ((\frac{x}{2})^{x} + (\frac{2}{x})^{x}) lo g_{2} x d x$ is equal to

(A)

(\frac{x}{2})^{x} + (\frac{2}{x})^{x} + C

(B)

(\frac{x}{2})^{x} - (\frac{2}{x})^{x} + C

(C)

(\frac{x}{2})^{x} lo g_{2} (\frac{x}{2}) + C

(D)

(\frac{2}{x})^{x} lo g_{2} (\frac{2}{x}) + C

JEE Main 2023 (10 April Shift 1)

LEVELJEE Main

If $I (x) = \int e^{s i n^{2} x} (cos x sin 2 x - sin x) d x$ and $I (0) = 1$ , then $I (\frac{π}{3})$ is equal to

(A)

- \frac{1}{2} e^{\frac{3}{4}}

(B)

\frac{1}{2} e^{\frac{3}{4}}

(C)

- e^{\frac{3}{4}}

(D)

e^{\frac{3}{4}}

JEE Main 2024 (06 April Shift 2)

LEVELJEE Main

If $\int \frac{1}{a ^{2} s i n ^{2} x + b ^{2} c o s ^{2} x} d x = \frac{1}{12} tan^{- 1} (3 tan x) + constant$ , then the maximum value of $a sin x + b cos x$ , is :

(A)

40

(B)

41

(C)

39

(D)

42

JEE Main 2024 (29 Jan Shift 1)

LEVELJEE Advanced

For $x \in (- \frac{π}{2}, \frac{π}{2})$ , if $y (x) = \int \frac{c s c x + s i n x}{c s c x s e c x + t a n x s i n ^{2} x} d x$ and $lim_{x \to (\frac{π}{2})^{-}} y (x) = 0$ then $y (\frac{π}{4})$ is equal to

(A)

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

(B)

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

(C)

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

(D)

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

JEE Main 2025 (January)

LEVELJEE Main

Let $I (x) = \int \frac{d x}{( x - 11 ) ^{\frac{1}{13}} ( x + 15 ) ^{\frac{15}{13}}} . I f I (37) - I (24) = \frac{1}{4} (\frac{1}{b ^{\frac{1}{13}}} - \frac{1}{c ^{\frac{1}{13}}})$ , b, $c \in N,$ then $3 (b + c)$ is equal to

JEE Main 2025 (January)

LEVELJEE Main

If $\int \frac{2 x ^{2} + 5 x + 9}{x ^{2} + x + 1} d x = x x^{2} + x + 1 + α x^{2} + x + 1 + β lo g_{e} ∣ x + \frac{1}{2} + x^{2} + x + 1 ∣ + C .$ where C is the constant of integration, then $α + 2 β$ is equal to

JEE Main 2025 (January)

LEVELJEE Main

If $\int e^{x} (\frac{x s i n ^{- 1} x}{1 - x ^{2}} + \frac{s i n ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} + \frac{x}{1 - x ^{2}}) d x = g (x) + C,$ where C is the constant of integration, then $g (\frac{1}{2})$ equals:

(A)

\frac{π}{4} \frac{e}{3}

(B)

\frac{π}{6} \frac{e}{3}

(C)

\frac{π}{4} \frac{e}{2}

(D)

\frac{π}{6} \frac{e}{2}

JEE Main 2026 (22 January Shift 1)

LEVELJEE Main

If $\int (sin x)^{- 11/2} (cos x)^{- 5/2} d x = - \frac{p _{1}}{q _{1}} (cot x)^{9/2} - \frac{p _{2}}{q _{2}} (cot x)^{5/2} - \frac{p _{3}}{q _{3}} (cot x)^{1/2} + \frac{p _{4}}{q _{4}} (cot x)^{- 3/2} + C$ , where $p_{i}$ and $q_{i}$ are positive integers with $g cd (p_{i}, q_{i}) = 1$ for $i = 1, 2, 3, 4$ and $C$ is the constant of integration, then $\frac{15 p _{1} p _{2} p _{3} p _{4}}{q _{1} q _{2} q _{3} q _{4}}$ is equal to

JEE Main 2026 (23 January Shift 2)

LEVELJEE Main

Let $I (x) = \int \frac{3 d x}{( 4 x + 6 ) ( 4 x ^{2} + 8 x + 3 )}$ and $I (0) = \frac{3}{4} + 20$ . If $I (\frac{1}{2}) = \frac{a 2}{b} + c$ , where $a, b, c \in N, g cd (a, b) = 1$ , then $a + b + c$ is equal to

JEE Main 2026 (24 January Shift 1)

LEVELJEE Main

Let $f (t) = \int (\frac{1 - s i n ( l o g _{e} t )}{1 - c o s ( l o g _{e} t )}) d t, t > 1$ . If $f (e^{π /2}) = - e^{π /2}$ and $f (e^{π /4}) = α e^{π /4}$ , then $α$ equals

(A)

1 + 2

(B)

- 1 - 2

(C)

- 1 - 22

(D)

- 1 + 2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫tanx−25tanx​dx=x+aln∣sinx−2cosx∣+k, then a is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫{1+(logx)2logx−1​}2dx is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫cosx−sinxdx​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let I=∫e4x+e2x+1ex​dx,J=∫e−4x+e−2x+1e−x​dx. Then, for an arbitrary constant C, the value of J−I equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluate ∫(x+1)3(x−1)ex​dx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluate ∫sinx−cosxsinx​dx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let In​=∫tannxdx,(n>1). If I4​+I6​=atan5x+bx5+C, where C is a constant of integration, then the ordered pair (a,b) is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫7−6x−x2​2x+5​dx=A7−6x−x2​+Bsin−1(4x+3​)+C (Where C is a constant of integration), then the ordered pair (A, B) is equal to :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫1+tanx+tan2xtanx​dx=x−A​K​tan−1(A​Ktanx+1​)+C, (C is a constant of integration), then the ordered pair (K, A) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫(x2−2x+10)2dx​=A(tan−1(3x−1​)+x2−2x+10f(x)​)+C, where C is a constant of integration, then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫5+7sinθ−2cos2θcosθ​dθ=Aloge​∣B(θ)∣+C where C is a constant of integration, then AB(θ)​ can be:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫(x2+x+1)2dx​=atan−1(3​2x+1​)+b(x2+x+12x+1​)+C,x>0 where C is the constant of integration, then the value of 9(3​a+b) is equal to.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫(x+1)2(x2+1)ex​dx=f(x)ex+C, Where C is constant, then dx3d3f​ at x=1 is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The integral ∫1+3​2​sin2x(1−3​1​)(cosx−sinx)​dx is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The integral ∫((2x​)x+(x2​)x)log2​xdx is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If I(x)=∫esin2x(cosxsin2x−sinx)dx and I(0)=1, then I(3π​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫a2sin2x+b2cos2x1​dx=121​tan−1(3tanx)+constant, then the maximum value of asinx+bcosx, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For x∈(−2π​,2π​), if y(x)=∫cscxsecx+tanxsin2xcscx+sinx​dx and limx→(2π​)−​y(x)=0 then y(4π​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let I(x)=∫(x−11)131​(x+15)1315​dx​.If I(37)−I(24)=41​(b131​1​−c131​1​) , b, c∈N, then 3(b+c) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫x2+x+1​2x2+5x+9​dx=xx2+x+1​+αx2+x+1​+β loge​∣x+21​+x2+x+1​∣+C. where C is the constant of integration, then α+2β is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫ex(1−x2​x sin−1x​+(1−x2)3/2sin−1x​+1−x2x​)dx=g(x)+C, where C is the constant of integration, then g(21​) equals:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let I(x)=∫(4x+6)(4x2+8x+3​)3dx​ and I(0)=43​​+20. If I(21​)=ba2​​+c, where a,b,c∈N,gcd(a,b)=1, then a+b+c is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(t)=∫(1−cos(loge​t)1−sin(loge​t)​)dt,t>1. If f(eπ/2)=−eπ/2 and f(eπ/4)=αeπ/4, then α equals

If $\int \frac{5 t a n x}{t a n x - 2} d x = x + a ln ∣ sin x - 2 cos x ∣ + k$ , then $a$ is equal to

$\int {\frac{l o g x - 1}{1 + ( l o g x ) ^{2}}}^{2} d x$ is equal to

$\int \frac{d x}{c o s x - s i n x}$ is equal to

Let $I = \int \frac{e ^{x}}{e ^{4 x} + e ^{2 x} + 1} d x, J = \int \frac{e ^{- x}}{e ^{- 4 x} + e ^{- 2 x} + 1} d x$ . Then, for an arbitrary constant $C$ , the value of $J - I$ equals

Evaluate $\int \frac{( x - 1 ) e ^{x}}{( x + 1 ) ^{3}} d x$

Evaluate $\int \frac{s i n x}{s i n x - c o s x} d x$

Let $I_{n} = \int tan^{n} x d x, (n > 1)$ . If $I_{4} + I_{6} = a tan^{5} x + b x^{5} + C$ , where C is a constant of integration, then the ordered pair $(a, b)$ is equal to:

If $\int \frac{2 x + 5}{7 - 6 x - x ^{2}} d x = A 7 - 6 x - x^{2} + B sin^{- 1} (\frac{x + 3}{4}) + C$ (Where C is a constant of integration), then the ordered pair (A, B) is equal to :-

If $\int \frac{t a n x}{1 + t a n x + t a n ^{2} x} d x = x - \frac{K}{A} tan^{- 1} (\frac{K t a n x + 1}{A}) + C$ , (C is a constant of integration), then the ordered pair (K, A) is equal to :

If $\int \frac{d x}{( x ^{2} - 2 x + 10 ) ^{2}} = A (tan^{- 1} (\frac{x - 1}{3}) + \frac{f ( x )}{x ^{2} - 2 x + 10}) + C$ , where $C$ is a constant of integration, then :

If $\int \frac{c o s θ}{5 + 7 s i n θ - 2 c o s ^{2} θ} d θ = A lo g_{e} ∣ B (θ) ∣ + C$ where $C$ is a constant of integration, then $\frac{B ( θ )}{A}$ can be:

If $\int \frac{d x}{( x ^{2} + x + 1 ) ^{2}} = a tan^{- 1} (\frac{2 x + 1}{3}) + b (\frac{2 x + 1}{x ^{2} + x + 1}) + C, x > 0$ where $C$ is the constant of integration, then the value of $9 (3 a + b)$ is equal to.

$\int \frac{( x ^{2} + 1 ) e ^{x}}{( x + 1 ) ^{2}} d x = f (x) e^{x} + C$ , Where $C$ is constant, then $\frac{d ^{3} f}{d x ^{3}}$ at $x = 1$ is equal to :

The integral $\int \frac{( 1 - \frac{1}{3} ) ( c o s x - s i n x )}{1 + \frac{2}{3} s i n 2 x} d x$ is equal to

The integral $\int ((\frac{x}{2})^{x} + (\frac{2}{x})^{x}) lo g_{2} x d x$ is equal to

If $I (x) = \int e^{s i n^{2} x} (cos x sin 2 x - sin x) d x$ and $I (0) = 1$ , then $I (\frac{π}{3})$ is equal to

If $\int \frac{1}{a ^{2} s i n ^{2} x + b ^{2} c o s ^{2} x} d x = \frac{1}{12} tan^{- 1} (3 tan x) + constant$ , then the maximum value of $a sin x + b cos x$ , is :

For $x \in (- \frac{π}{2}, \frac{π}{2})$ , if $y (x) = \int \frac{c s c x + s i n x}{c s c x s e c x + t a n x s i n ^{2} x} d x$ and $lim_{x \to (\frac{π}{2})^{-}} y (x) = 0$ then $y (\frac{π}{4})$ is equal to

Let $I (x) = \int \frac{d x}{( x - 11 ) ^{\frac{1}{13}} ( x + 15 ) ^{\frac{15}{13}}} . I f I (37) - I (24) = \frac{1}{4} (\frac{1}{b ^{\frac{1}{13}}} - \frac{1}{c ^{\frac{1}{13}}})$ , b, $c \in N,$ then $3 (b + c)$ is equal to

If $\int \frac{2 x ^{2} + 5 x + 9}{x ^{2} + x + 1} d x = x x^{2} + x + 1 + α x^{2} + x + 1 + β lo g_{e} ∣ x + \frac{1}{2} + x^{2} + x + 1 ∣ + C .$ where C is the constant of integration, then $α + 2 β$ is equal to

If $\int e^{x} (\frac{x s i n ^{- 1} x}{1 - x ^{2}} + \frac{s i n ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} + \frac{x}{1 - x ^{2}}) d x = g (x) + C,$ where C is the constant of integration, then $g (\frac{1}{2})$ equals:

Let $I (x) = \int \frac{3 d x}{( 4 x + 6 ) ( 4 x ^{2} + 8 x + 3 )}$ and $I (0) = \frac{3}{4} + 20$ . If $I (\frac{1}{2}) = \frac{a 2}{b} + c$ , where $a, b, c \in N, g cd (a, b) = 1$ , then $a + b + c$ is equal to

Let $f (t) = \int (\frac{1 - s i n ( l o g _{e} t )}{1 - c o s ( l o g _{e} t )}) d t, t > 1$ . If $f (e^{π /2}) = - e^{π /2}$ and $f (e^{π /4}) = α e^{π /4}$ , then $α$ equals