datavorous/entrance-exam-dataset huggingface dataset:
The integral $\int_{\pi / 6}^{\pi / 4} \frac{\mathrm{d} x}{\sin 2 x\left(\tan ^{5} x+\cot ^{5} x\right)}$ equals:

Options:
A $\frac{1}{20} \tan ^{-1}\left(\frac{1}{9 \sqrt{3}}\right)$
B $\frac{1}{10}\left(\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{9 \sqrt{3}}\right)\right)$
C $\frac{\pi}{40}$
D $\frac{1}{5}\left(\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{3 \sqrt{3}}\right)\right)$

The correct answer is:
$\frac{1}{10}\left(\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{9 \sqrt{3}}\right)\right)$

Solution:
$\mathrm{I}=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{d x}{\sin 2 x\left(\tan ^{5} x+\cot ^{5} x\right)}$

$\begin{array}{l}

=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\tan ^{5} x \cdot \sec ^{2} x}{2 \frac{\sin x}{\cos x}\left(\left(\tan ^{5} x\right)^{2}+1\right)} \\

=\frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{4} \tan ^{4} x \cdot \sec ^{2} x}{\left(\tan ^{5} x\right)^{2}+1} d x

\end{array}$

Let $\tan ^{4} x=t$

$5 \tan ^{4} x \cdot \sec ^{2} x d x=d t$

When $x \rightarrow \frac{\pi}{4}$ then $t \rightarrow 1$

and $x \rightarrow \frac{\pi}{6}$ then $t \rightarrow\left(\frac{1}{\sqrt{3}}\right)^{5}$

$\begin{aligned}

\therefore \quad & \mathrm{I}=\frac{1}{10} \int_{\left(\frac{1}{\sqrt{3}}\right)^{5}}^{1} \frac{d t}{t^{2}+1} \\

&=\frac{1}{10}\left(\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{9 \sqrt{3}}\right)\right)

\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2019 (11 Jan Shift 2)
The integral $\int_{7 \pi / 4}^{7 \pi / 3} \sqrt{\tan ^2 x} d x$ is equal to :

Options:
A
$\log 2 \sqrt{2}$

B
$\log 2$

C
$2 \log 2$

D
$\log \sqrt{2}$

The correct answer is:
$\log \sqrt{2}$

Solution:
Let $\mathrm{I}=\int_{7 \pi / 4}^{7 \pi / 3} \sqrt{\tan ^2 x} d x$
$$
=\int_{7 \pi / 4}^{7 \pi / 3} \tan x d x=-\left.\log \cos x\right|_{7 \pi / 4} ^{7 \pi / 3}
$$
$=-\left[\log \cos \frac{7 \pi}{3}-\log \cos \frac{7 \pi}{4}\right]$
$=\log \cos \frac{7 \pi}{4}-\log \cos \frac{7 \pi}{3}$
$=\log \left[\frac{\cos \frac{7 \pi}{4}}{\cos \frac{7 \pi}{3}}\right]=\log \left[\frac{\cos \left(2 \pi-\frac{\pi}{4}\right)}{\cos \left(2 \pi+\frac{\pi}{3}\right)}\right]$
$=\log \left(\frac{\cos \frac{\pi}{4}}{\cos \frac{\pi}{3}}\right)=\log \left(\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\right)$
$=\log \left(\frac{2}{\sqrt{2}}\right)=\log \sqrt{2}$.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2013 (22 Apr Online)
The integral ∫π12π48cos2xtanx+cotx3dx equals

Options:
A 13256
B 1564
C 1332
D 15128

The correct answer is:
15128

Solution:
Given

∫π12π48cos2xtanx+cotx3dx=∫π12π48cos2xsinxcosx+cosxsinx3dx

=∫π12π4cos2x1sin2x3=∫π12π4cos2x·sin2x·sin22xdx

=14∫π12π4sin4x·(1−cos4x)dx

=14∫π12π4sin4x-18∫π12π4sin8x

=−116cos4xπ12π4+18×8cos8xπ12π4

=−116−1−12+1641+12

=15128.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2017 (08 Apr Online)
The integral $\int_{1 / 4}^{3 / 4} \cos \left(2 \cot ^{-1} \sqrt{\frac{1-x}{1+x}}\right) d x$ is equal to

Options:
A $1 / 2$
B $-1 / 2$
C $-1 / 4$
D $1 / 4$

The correct answer is:
$-1 / 4$

Solution:
$\begin{aligned} & I=\int_{1 / 4}^{3 / 4} \cos \left(2 \cot ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right) d x\right) \\ & \int_{1 / 4}^{3 / 4} \cos \left(2\left(\tan ^{-1} \sqrt{\frac{1+x}{1+x}}\right)\right) d x \\ & \left.\int_{1 / 4}^{3 / 4} \frac{1-\tan ^2\left(\tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right)}{1+\tan ^2\left(\tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right.}\right) d x\end{aligned}$
$\begin{aligned} & =\int_{1 / 4}^{3 / 4} \frac{1-\left(\frac{1+x}{1-x}\right)}{1+\left(\frac{1+x}{1-x}\right)} d x=\int_{1 / 4}^{3 / 4} \frac{-2 x}{2} d x \\ & =\int_{1 / 4}^{3 / 4}(-x) d x=-\left(\frac{x^2}{2}\right)_{1 / 4}^{3 / 4} \\ & =-\frac{1}{2}\left[\frac{9}{16}-\frac{1}{16}\right] \\ & =-\frac{1}{4}\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2024 (09 Apr Shift 2)
The integral 24π∫022-x2dx2+x24+x4 is equal to ______.

Options:

The correct answer is:
3

Solution:
Let, I=24π∫022-x22+x24+x4dx

=24π∫022-x2x22x+x4x2+x2dx

=24π∫022x2-12x+x2x+x2-4dx

Now let 2x+x=t, -2x2+1dx=dt

I=-24π∫∞22dttt2-4=-12π∫∞222tdtt2t2-4

Let t2-4=z2, 2tdt=2zdz

I=-12π∫∞22zdzzz2+4=-24π∫∞2dzz2+4=-24π12tan-1z2∞2

=-24ππ8-π4=3

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (26 Jun Shift 2)
The integral ∫coslnxdx, is equal to

Options:
A x2coslnx-sinlnx+C
B xcoslnx-sinlnx+C
C xcoslnx+sinlnx+C
D x2coslnx+sinlnx+C

The correct answer is:
x2coslnx+sinlnx+C

Solution:
Given, I=∫coslnxdx
Let, lnx=t⇒x=et⇒dx=etdt

∴I=∫etcostdt
∫etcostdt=12∫etcost+sint+cost-sintdt
∫etcost dt=12etcost+sint+C, where C is the constant of integration.
=x2coslnx+sinlnx+C.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2019 (12 Jan Shift 1)
The integral ∫dx1+xx-x2 is equal to

Options:
A -21+x1-x+c
B -1-x1+x+c
C -2 1-x1+x+c
D 1+x1-x+c

The correct answer is:
-2 1-x1+x+c

Solution:
I=∫dx1+xx1-x

Put, 1+x=t ⇒12xdx=dt

And, 1-x=1-t-12=2t-t2
∴ I= ∫2dtt2t-t2
Again put, t=1z⇒dt=-1z2dz
⇒I=2∫-1z2 dz1z 2z-1z2=2∫-dz2z-1=-22z-1+c

Where c is arbitrary constant.

=-22t-1 +c
=-22-tt+c
=-2 1-x1+x+c

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2016 (10 Apr Online)
The integral ∫dxx+134x-254, is equal to

Options:
A 4x-2x+114+c
B -43 x+1x-214+c
C 4x+1x-214+c
D -43x-2x+114+c

The correct answer is:
-43 x+1x-214+c

Solution:
∫dxx+134x-254
=∫dxx+12x-2x+154
Put, x-2x+1=t⇒3x+12dx=dt
∴I=-431t14+c
=-43x+1x-214+c, c being the constant of integration.

Topic: Mathematics, Indefinite Integration, JEE Main
The integral ∫dxx+134x-254, is equal to

Options:
A 4x-2x+114+c
B -43 x+1x-214+c
C 4x+1x-214+c
D -43x-2x+114+c

The correct answer is:
-43 x+1x-214+c

Solution:
∫dxx+134x-254
=∫dxx+12x-2x+154
Put, x-2x+1=t⇒3x+12dx=dt
∴I=-431t14+c
=-43x+1x-214+c, c being the constant of integration.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2015 (10 Apr Online)
The integral ∫dxx2x4+134 equals to

Options:
A -x4+1x414+c
B x4+1x414+c
C x4+114+c
D -x4+114+c

The correct answer is:
-x4+1x414+c

Solution:
I=∫dxx2x4+134
=∫dxx51+1x434
Put, 1+1x4=t⇒-4x5dx=dt
∴ I=14∫-dtt34
= - 1 4 × ∫ t - 3 4 dt =-14t1414+c
=-1+1x414+c, where c is the constant of integration.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2015 (04 Apr)
The integral ∫dx(x+4)87(x-3)67 is equal to: (where C is a constant of integration)

Options:
A x-3x+417+C
B x-3x+4-17+C
C 12x-3x+437+C
D -113x-3x+4-137+C

The correct answer is:
x-3x+417+C

Solution:
⇒∫x-3x+4-671x+42dx ...i

Let x-3x+4=t7 ,

7x+42dx=7t6dt ... ii

From i from ii,

∫t-6t6dt=t+C

Topic: Mathematics, Indefinite Integration, JEE Main
The integral ∫dx(x+4)87(x-3)67 is equal to: (where C is a constant of integration)

Options:
A x-3x+417+C
B x-3x+4-17+C
C 12x-3x+437+C
D -113x-3x+4-137+C

The correct answer is:
x-3x+417+C

Solution:
⇒∫x-3x+4-671x+42dx ...i

Let x-3x+4=t7 ,

7x+42dx=7t6dt ... ii

From i from ii,

∫t-6t6dt=t+C

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2020 (09 Jan Shift 1)
The integral ∫e3loge2x+5e2loge2xe4logex+5e3logex-7e2logex dx, x>0, is equal to
(where c is a constant of integration)

Options:
A logex2+5x-7+c
B 4logex2+5x-7+c
C 14logex2+5x-7+c
D logex2+5x-7+c

The correct answer is:
4logex2+5x-7+c

Solution:
∫e3loge2x+5e2loge2xe4logex+5e3logex-7e2logexdx,x>0

=∫2x3+52x2x4+5x3-7x2dx=∫4x22x+5x2x2+5x-7dx

=4∫dx2+5x-7x2+5x-7=4logex2+5x-7+c

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2021 (25 Feb Shift 2)
The integral Ι=∫ex1+sin⁡x1+cos⁡xdx=exfx+C (where, C is the constant of integration). Then, the range of y=fx (for all x in the domain of fx) is

Options:
A -1,1
B -∞,∞
C -1,1
D 0,∞

The correct answer is:
-∞,∞

Solution:
The given integral Ι=∫ex1+2sinx2cosx22cos2x2dx
=∫ex12sec2x2+tanx2dx
=ex⋅tanx2+C ∵ ∫exfx+f'xdx=exfx+C
Thus, fx=tanx2 whose range is -∞,∞

Topic: Mathematics, Indefinite Integration, JEE Main
The integral I=â«0100Ïtan-1xdx, (where, . represents the greatest integer function) has the value KÏ+tanâ¡p, then the value of K+p is equal to

Options:
A 101
B 99
C 100π
D 99π

The correct answer is:
99

Solution:
The given integral is I=∫0tan1⁡0dx+∫tan⁡1100π⁡1dx
as tan-1x⁡∈1,π2, ∀x>tan⁡1
⇒I=100π-tan⁡1
=100π+tan⁡-1
∴k=100, p=-1
∴k+p=99

Topic: Mathematics, Definite Integration, JEE Main
The integral I=â«1xâsecâ¡x-lnâ¡xsinâ¡xdx simplifies to (where, c is the constant of integration)

Options:
A ln⁡xsin⁡x+c
B ln⁡xcos⁡x+c
C xln⁡sin⁡x+c
D xln⁡cos⁡x+c

The correct answer is:
ln⁡xcos⁡x+c

Solution:
Given I=∫cos⁡x⋅1x+-sin⁡xln⁡xdx
Since,∫f⋅g'+f'gdx
=f⋅g+c
⇒I=cos⁡xln⁡x+c

Topic: Mathematics, Indefinite Integration, JEE Main
The integral I=∫22x+xdx=λ.22x+C (where, C is the constant of integration). Then the value of λ is equal to

Options:
A 1ln⁡4
B 1ln⁡22
C 1ln⁡2
D 1ln⁡42

The correct answer is:
1ln⁡2

Solution:
Given, I=â«22x.2xdx
Let, 22x=tâ22x.2x.lnâ¡22dx=dt
â22x.2xdx=dtlnâ¡22
Thus, I=â«dtlnâ¡22=tlnâ¡22+C
=22xlnâ¡22+C
â´Î»=1lnâ¡22âÎ»=llnâ¡2

Topic: Mathematics, Indefinite Integration, JEE Main
The integral I=∫2sin⁡x3+sin⁡2xdx simplifies to (where, C is the constant of integration)

Options:
A ln⁡2+sin⁡x-cos⁡x2-sin⁡x+cos⁡x-tan-1⁡sin⁡x+cos⁡x+C
B ln⁡(sin⁡x)+sin⁡2x+C
C sin⁡2x-ln⁡(cos⁡x)+C
D 14ln⁡2+sin⁡x-cos⁡x2-sin⁡x+cos⁡x-12tan-1⁡sin⁡x+cos⁡x2+C

The correct answer is:
14ln⁡2+sin⁡x-cos⁡x2-sin⁡x+cos⁡x-12tan-1⁡sin⁡x+cos⁡x2+C

Solution:
The given integral can be rewritten as,

I=∫sin⁡x+cos⁡x3+sin⁡2xdx+∫sin⁡x-cos⁡x3+sin⁡2xdx
Let sin⁡x-cos⁡x=t and sin⁡x+cos⁡x=u in 1st and 2nd integral respectively.
∴I=∫dt4-t2-∫du2+u2
=14ln⁡2+t2-t-12tan-1⁡u2+C
Putting values of t and u, we get,

I=14ln⁡2+sinx-cosx2-sinx+cosx-12tan-1⁡sinx+cosx2+C

Topic: Mathematics, Indefinite Integration, JEE Main
The integral I=∫π4π2sin6⁡xdx satisfies

Options:
A I>π2
B I>π4
C I∈π4,π2
D I∈π32,π4

The correct answer is:
I∈π32,π4

Solution:
∫π4π2sin6⁡π4dx<∫π4π2sin6⁡xdx<∫π4π2sin6⁡π2dx
⇒18∫π4π2dx<I<∫π4π2dx
⇒18⋅π4<I<π4
or I∈π32,π4

Topic: Mathematics, Definite Integration, JEE Main
The integral I=∫ee+11+x21+x3dx satisfies

Options:
A I>2
B I>e
C I<0
D I<1

The correct answer is:
I<1

Solution:
x∈e,e+1
⇒x2<x3
⇒1+x21+x3<1
⇒∫ee+11+x21+x3dx<∫ee+11dx
⇒I<xee+1
⇒I<1

Topic: Mathematics, Definite Integration, JEE Main
The integral I=∫eesin⁡x+sin⁡xcos⁡xdx simplifies to (where, c is the constant of integration)

Options:
A esin⁡x+c
B esin⁡x+cos⁡x+c
C eecos⁡x+c
D eesin⁡x+c

The correct answer is:
eesin⁡x+c

Solution:
The given integral is I=∫eesin⁡xesin⁡xcos⁡xdx
Let eesin⁡x=t
⇒eesin⁡xesin⁡xcos⁡xdx=dt
∴I=∫1dt=t+c
⇒I=eesin⁡x+c

Topic: Mathematics, Indefinite Integration, JEE Main
The integral I=â«excosâ¡exxdx=fx+c (where, c is the constant of integration) and flnâ¡Ï42=2. Then, the number of solutions of fx=2e âxâR-0 is equal to

Options:

The correct answer is:
0

Solution:
Given integral is
I=∫excos⁡exxdx
Let ex=t
⇒ex2xdx=dt
∴I=2∫cos⁡tdt
=2sin⁡t+c
=2sin⁡ex+c
∴fx=2sin⁡ex
which has range =-2,2
Hence, fx=2e has no solution

Topic: Mathematics, Indefinite Integration, JEE Main
The integral I=∫sec3⁡xtan3⁡xdx is equal to (where, C is the constant of integration)

Options:
A sec5⁡x-sec3⁡x+C
B sec5⁡x5-sec3⁡x+C
C sec5⁡x5-sec3⁡x3+C
D sec5⁡x5-tan-1⁡x+C

The correct answer is:
sec5⁡x5-sec3⁡x3+C

Solution:
Let sec⁡x=t⇒sec⁡xtan⁡xdx=dt
∴I=∫sec2⁡xtan2⁡xsecxtanxdx
=∫t2t2-1dt
=∫t4-t2dt
=t55-t33+C
I=sec5⁡x5-sec3⁡x3+C

Topic: Mathematics, Indefinite Integration, JEE Main
The integral I=∫sin⁡2θ1+cos2⁡θ2sin2⁡θdθ simplifies to (where, c is the integration constant)

Options:
A ln⁡sin⁡θ+cos⁡θ+c
B 2ln⁡sin⁡θ-sin2⁡θ2+c
C ln⁡sin⁡θ-sin2⁡θ+c
D ln⁡cos⁡θ+cos2⁡θ+c

The correct answer is:
2ln⁡sin⁡θ-sin2⁡θ2+c

Solution:
I=∫cos⁡θ1+cos2⁡θsin⁡θdθ
Let sin⁡θ=t
⇒cos⁡θdθ=dt
⇒I=∫2-t2tdt
=2ln⁡t-t22+c
=2ln⁡sin⁡θ-sin2⁡θ2+c

Topic: Mathematics, Indefinite Integration, JEE Main
The integral I=∫sin3⁡θcos⁡θ1+sin2⁡θ2dθ simplifies to (where, c is the constant of integration)

Options:
A 12ln⁡sin⁡θ+11+sin2⁡θ+c
B 12lnâ¡1+sin2â¡Î¸+11+sin2â¡Î¸+c
C lnâ¡sinâ¡Î¸+11+sin2â¡Î¸+c
D lnâ¡sin2â¡Î¸+1+1sin2â¡Î¸+2+c

The correct answer is:
12ln⁡1+sin2⁡θ+11+sin2⁡θ+c

Solution:
Let 1+sin2⁡θ=t
⇒2sin⁡θcos⁡θdθ=dt
∴I=12∫t-1t2dt
=12∫1t-1t2dt
=12ln⁡t+1t+c
=12ln⁡1+sin2⁡θ+11+sin2⁡θ+c

Topic: Mathematics, Indefinite Integration, JEE Main
The integral I=â«xex2sinâ¡x2+cosâ¡x2dx=fx+c, (where, c is the constant of integration). Then, fx can be

Options:
A exsin⁡x2
B ex2sin⁡x
C ex2sin⁡x22
D 12ex2sin⁡x2

The correct answer is:
12ex2sin⁡x2

Solution:
Let, x2=tâ2xdx=dt
â´I=12â«etsinâ¡t+cosâ¡tdt
=12etâsinâ¡t+c
=12ex2sinâ¡x2+c
As,ââ« e x f x + f â² x dx= e x âf x +c

Topic: Mathematics, Indefinite Integration, JEE Main
The integral \( \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec ^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x d x \) is equal to

Options:
A \( 3^{\frac{7}{6}}-3^{\frac{5}{6}} \)
B \( 3^{\frac{4}{3}}-3^{\frac{1}{3}} \)
C \( 3^{\frac{5}{6}}-3^{\frac{2}{3}} \)
D \( 3^{\frac{5}{3}}-3^{\frac{1}{3}} \)

The correct answer is:
\( 3^{\frac{7}{6}}-3^{\frac{5}{6}} \)

Solution:
I=∫π6π3sec23xcosec43x dx

=∫π6π3sec2xtan4/3xdx

Let tanx=t, sec2xdx=dt

⇒I=∫133 dtt43

=t-43+1-43+1133

=-31316- 316

=376-356.

Topic: Mathematics, Definite Integration, JEE Main
The integral of $\frac{x^2-x}{x^3-x^2+x-1}$ w.r.t. $x$ is

Options:
A
$\frac{1}{2} \log \left(x^2+1+c\right)$

B
$\frac{1}{2} \log \left|x^2-1\right|+c$

C
$\log \left(x^2+1+c\right)$

D
$\log \left|x^2-1\right|+c$

The correct answer is:
$\frac{1}{2} \log \left(x^2+1+c\right)$

Solution:
Let $I=\int \frac{x^2-x}{x^3-x^2+x-1} d x$
$$
=\int \frac{x(x-1)}{x^2(x-1)+(x-1)} d x=\int \frac{x d x}{x^2+1}
$$
$$
\left.=\frac{1}{2} \int \frac{2 x d x}{\left(x^2+1\right.}\right)
$$
Let $x^2+1=t \Rightarrow 2 x d x=d t$
$$
\begin{aligned}
& \therefore I=\frac{1}{2} \int \frac{d t}{t}=\frac{1}{2} \log t+c \\
& =\frac{1}{2} \log \left(x^2+1\right)+c
\end{aligned}
$$
where ' $c$ ' is the constant of integration.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2012 (12 May Online)
The integral $\int \frac{\sec ^{2} x}{(\sec x+\tan x)^{\frac{9}{2}}} d x$ equals (for some arbitrary constant $\mathrm{K}$ )

Options:
A $-\frac{1}{(\sec x+\tan x)^{\frac{11}{2}}}\left\{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$
B $\frac{1}{(\sec x+\tan x)^{\frac{11}{2}}}\left\{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$
C $-\frac{1}{(\sec x+\tan x)^{\frac{11}{2}}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$
D $\frac{1}{(\sec x+\tan x)^{\frac{11}{2}}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$

The correct answer is:
$-\frac{1}{(\sec x+\tan x)^{\frac{11}{2}}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$

Solution:
$\begin{array}{l}

I=\int \frac{\sec ^{2} x}{(\sec x+\tan x)^{9 / 2}} d x \\

\text { Let } \sec x+\tan x=t \Rightarrow \sec x-\tan x=\frac{1}{t} \\

\Rightarrow \sec x=\frac{1}{2}\left(t+\frac{1}{t}\right) \text { and } \sec x(\sec x+\tan x) d x=d t \\

\Rightarrow \sec x d x=\frac{d t}{t} \\

\therefore \quad I=\frac{1}{2} \int \frac{\left(t+\frac{1}{t}\right) d t}{t^{9 / 2} \cdot t}=\frac{1}{2} \int\left(t^{-9 / 2}+t^{-13 / 2}\right) d t \\

=\frac{-1}{7} t^{-7 / 2}-\frac{1}{11} t^{-1 / 2}+K \\

=-\frac{1}{7 t^{7 / 2}}-\frac{1}{11 t^{11 / 2}}+K=-\frac{1}{t^{11 / 2}}\left(\frac{1}{11}+\frac{t^{2}}{7}\right)+K \\

=\frac{-1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K

\end{array}$

Topic: Mathematics, Indefinite Integration, JEE Main
$$
\text { The integral } \int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^3 x+\cos ^3 x\right)^2} d x \text { is equal to: }
$$

Options:
A
$\frac{1}{\left(1+\cot ^3 x\right.}+c$

B
$\left.-\frac{1}{3\left(1+\tan ^3 x\right.}+c\right)$

C
$\frac{\sin ^3 x}{\left(1+\cos ^3 x\right.}+c$

D
$-\frac{\cos ^3 x}{3\left(1+\sin ^3 x\right.}+c$

The correct answer is:
$\left.-\frac{1}{3\left(1+\tan ^3 x\right.}+c\right)$

Solution:
Let $I=\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^3 x+\cos ^3 x\right)^2} d x$
$$
\begin{aligned}
&\mathrm{I}=\int\left(\frac{\sin x \cdot \cos x}{\sin ^3 x+\cos ^3 x}\right)^2 d x \\
&\mathrm{I}=\int\left(\frac{\sin x \cdot \cos x}{\cos ^3 x\left(1+\tan ^3 x\right)}\right)^2 d x \\
&=\int\left(\frac{\sin x \cdot \sec ^2 x}{\left(1+\tan ^3 x\right)}\right)^2 d x \\
&\text { Put } 1+\tan ^3 x=\mathrm{t} \\
&d t=3 \tan ^2 x \sec ^2 x d x \text { or } d x \\
&=\frac{d t}{3 \tan ^2 x \sec ^2 x} \\
&\therefore \mathrm{I}=\int \frac{\sin ^2 x \cdot \sec ^4 x}{t^2} \times \frac{d t}{3 \tan ^2 x \sec ^2 x}
\end{aligned}
$$
$$
\begin{aligned}
&\mathrm{I}=\frac{1}{3} \int \frac{\sin ^2 x \cdot \sec ^4 x}{t^2} \times \frac{d t}{\frac{\sin ^2 x}{\cos ^2 x} \times \sec ^2 x} \\
&=\frac{1}{3} \int \frac{\sin ^2 x \cdot \sec ^4 x}{t^2} \times \frac{d t}{\sin ^2 x \sec ^4 x} \\
&\therefore \mathrm{I}=\frac{1}{3} \int \frac{d t}{t^2}=\frac{1}{3} \int t^{-2} d t \\
&\mathrm{I}=\frac{1}{3}\left[\frac{t^{-2+1}}{-2+1}\right]+c=\frac{-1}{3}\left[\frac{1}{t}\right]+c \\
&\text { or } \mathrm{I}=-\frac{1}{3\left(1+\tan ^3 x\right)}+c
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, JEE Main
$$
\text { The integral } \int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^3 x+\cos ^3 x\right)^2} d x \text { is equal to: }
$$

Options:
A
$\frac{1}{\left(1+\cot ^3 x\right.}+c$

B
$\left.-\frac{1}{3\left(1+\tan ^3 x\right.}+c\right)$

C
$\frac{\sin ^3 x}{\left(1+\cos ^3 x\right.}+c$

D
$-\frac{\cos ^3 x}{3\left(1+\sin ^3 x\right.}+c$

The correct answer is:
$\left.-\frac{1}{3\left(1+\tan ^3 x\right.}+c\right)$

Solution:
Let $I=\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^3 x+\cos ^3 x\right)^2} d x$
$$
\begin{aligned}
&\mathrm{I}=\int\left(\frac{\sin x \cdot \cos x}{\sin ^3 x+\cos ^3 x}\right)^2 d x \\
&\mathrm{I}=\int\left(\frac{\sin x \cdot \cos x}{\cos ^3 x\left(1+\tan ^3 x\right)}\right)^2 d x \\
&=\int\left(\frac{\sin x \cdot \sec ^2 x}{\left(1+\tan ^3 x\right)}\right)^2 d x \\
&\text { Put } 1+\tan ^3 x=\mathrm{t} \\
&d t=3 \tan ^2 x \sec ^2 x d x \text { or } d x \\
&=\frac{d t}{3 \tan ^2 x \sec ^2 x} \\
&\therefore \mathrm{I}=\int \frac{\sin ^2 x \cdot \sec ^4 x}{t^2} \times \frac{d t}{3 \tan ^2 x \sec ^2 x}
\end{aligned}
$$
$$
\begin{aligned}
&\mathrm{I}=\frac{1}{3} \int \frac{\sin ^2 x \cdot \sec ^4 x}{t^2} \times \frac{d t}{\frac{\sin ^2 x}{\cos ^2 x} \times \sec ^2 x} \\
&=\frac{1}{3} \int \frac{\sin ^2 x \cdot \sec ^4 x}{t^2} \times \frac{d t}{\sin ^2 x \sec ^4 x} \\
&\therefore \mathrm{I}=\frac{1}{3} \int \frac{d t}{t^2}=\frac{1}{3} \int t^{-2} d t \\
&\mathrm{I}=\frac{1}{3}\left[\frac{t^{-2+1}}{-2+1}\right]+c=\frac{-1}{3}\left[\frac{1}{t}\right]+c \\
&\text { or } \mathrm{I}=-\frac{1}{3\left(1+\tan ^3 x\right)}+c
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2014 (12 Apr Online)
The integral Ι=∫sinx2+2x2cosx2dx=xHx+C, (where C is the constant of integration). If the range of H(x) is a,b, then the value of a+2b is equal to

Options:

The correct answer is:
1

Solution:
Given integral Ι=∫1⋅sinx2+x⋅2xcosx2dx
∵∫f'g+f⋅g'dx=∫fg'dx=f⋅g+C
=x⋅sinx2+C
∴ Hx=sinx2, whose range is -1,1
∴ a=-1, b=1⇒a+2b=1

Topic: Mathematics, Indefinite Integration, JEE Main
The integral ∫sin2xcos2xsin5x+cos3xsin2x+sin3xcos2x+cos5x2dx, is equal to
(where C is the constant of integration).

Options:
A -11+cot3x+C
B 131+tan3x+C
C -131+tan3x+C
D 11+cot3x+C

The correct answer is:
-131+tan3x+C

Solution:
∫sin2xcos2xsin5x+cos3xsin2x+sin3xcos2x+cos5x2dx
=∫sin2xcos2xsin3xsin2x+cos2x+cos3x cos2x+sin2x2dx
=∫sin2xcos2xsin3x+cos3x2dx
=∫sin2xcos2xcos3x2tan3x+12dx
=∫tan2xsec2xtan3x+12dx
Put, tan3x+1=t
⇒dt=3tan2xsec2xdx
=∫dt3t2=-13t+C, where C is the constant of integration.
= -13tan3x+1+C

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2018 (08 Apr)
The integral value of a, for which the quadratic expression ax2+a-2x-2 is negative for exactly two integral values of x, is

Options:

The correct answer is:
1

Solution:
Let, fx=ax2+a-2x-2
∴f0=-2 and f-1=0
Since, the quadratic expression is negative for exactly two integral values
⇒f1<0 and f2≥0
⇒a+a-2-2<0 and 4a+2a-4-2≥0
⇒a<2 and a≥1
⇒a∈1,2
Hence, the integral value of a=1

Topic: Mathematics, Quadratic Equation, JEE Main
The integral value of \( \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{x}{1+\sin x} d x= \)

Options:
A \( \pi(\sqrt{2}+1) \)
B \( 2 \pi(\sqrt{2}-1) \)
C \( 2 \pi(\sqrt{2}+1) \)
D \( \frac{\pi}{\sqrt{2}+1} \)

The correct answer is:
\( \frac{\pi}{\sqrt{2}+1} \)

Solution:
Let I=∫π43π4x1+sin⁡xdx ....(i)
I=∫π43π43π4+π4-x1+sin⁡3π4+π4-x
= ∫π43π4π-xdx1+sin⁡x ....(ii)
∵ ∫abfxdx=∫abfa+b-xdx
By adding Equations (i) and (ii), we get
2I= ∫π43π4π dx1+sin⁡x
⇒ 2I=π∫π43π41-sin⁡x1+sin⁡x 1-sin⁡xdx
⇒2I=π∫π43π41-sin⁡xcos2⁡xdx
⇒2I=π∫π43π4sec2⁡x-sec⁡xtan⁡xdx
⇒2I=πtan⁡x-sec⁡xπ43π4
⇒2I=π-1--2-1-2
⇒2I=π-1+2-1+2
⇒2I=π-2+22
∴I=π2-1
=π2-12+12+1
=π2+1

Topic: Mathematics, Definite Integration, JEE Main
The integral value of m for which the quadratic equation 2m-3x2-4x+2m-3=0 has both the roots negative is given by

Options:
A 1
B 2
C 3
D 5

The correct answer is:
1

Solution:
âµ Both roots are negative
â´D>0, sum of roots â-ve, product of roots â+ve
D>0â16-42m-32>0â4-2m-32>0
2-2m+32+2m-3>0â-2m+52m-1>0
â2m-52m-1<0â12<m<52
Î±+Î²<0â42m-3<0âm<32
Î±Î²>0â1>0 (always true)
Taking intersection, we get, mâ12,32
Integral value of m=1

Topic: Mathematics, Quadratic Equation, JEE Main
The integral ∫x2x+2xxlog2x dx is equal to

Options:
A x2x+2xx+C
B x2x-2xx+C
C x2xlog2x2+C
D x2xlog22x+C

The correct answer is:
x2x+2xx+C

Solution:
Given,

I=∫x2x+2xxlog2x dx

Now, Let x2x=t

⇒ xlog2x2=log2t

⇒ xlogex2·log2e=loget·log2e

On differentiating both sides we get :
logex2+x·2x·12=1tdtdx

⇒logex2+1=1tdtdx

Then solution is not possible as there is no proper substitution.

Note: This question was bonus in Jee Mains 2023 April session.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The integral $\int x \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) d x(x>0)$ is equal to:

Options:
A
$-x+\left(1+x^2\right) \tan ^{-1} x+c$

B
$x-\left(1+x^2\right) \cot ^{-1} x+c$

C
$-x+\left(1+x^2\right) \cot ^{-1} x+c$

D
$x-\left(1+x^2\right) \tan ^{-1} x+c$

The correct answer is:
$-x+\left(1+x^2\right) \tan ^{-1} x+c$

Solution:
Let $\mathrm{I}=\int x \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) d x$
$$
\therefore \mathrm{I}=2 \int_{\text {II }} x \cdot \tan ^{-1} x d x
$$
Applying Integration by parts
$$
\begin{array}{r}
\mathrm{I}=2\left[\tan ^{-1} x \int x d x-\int\left(\frac{d}{d x}\left(\tan ^{-1} x\right) \int x d x\right) d x\right] \\
\mathrm{I}=2\left[\frac{x^2}{2} \tan ^{-1} x-\int \frac{1}{1+x^2} \times \frac{x^2}{2} d x\right]+c \\
\mathrm{I}=2\left[\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int \frac{x^2+1-1}{x^2+1} d x\right]+c \\
\mathrm{I}=2\left[\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int \frac{x^2+1}{x^2+1} d x+\frac{1}{2} \int \frac{1}{1+x^2} d x\right]+c
\end{array}
$$
$$
\begin{aligned}
\mathrm{I} &=2\left[\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int 1 . d x+\frac{1}{2} \tan ^{-1} x\right]+c \\
\mathrm{I} &=2\left[\frac{x^2}{2} \tan ^{-1} x-\frac{x}{2}+\frac{1}{2} \tan ^{-1} x\right]+c \\
\mathrm{I} &=x^2 \tan ^{-1} x+\tan ^{-1} x-x+c \\
& \text { or } I=-x+\left(x^2+1\right) \tan ^{-1} x+c
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2014 (11 Apr Online)
The integral â« x cos - 1 1 - x 2 1 + x 2 dx; x > 0 is equal to

Options:
A - x + 1 + x 2 cot - 1 x + c
B x - 1 + x 2 cot - 1 x + c
C x - 1 + x 2 tan - 1 x + c
D - x + 1 + x 2 tan - 1 x + c

The correct answer is:
- x + 1 + x 2 tan - 1 x + c

Solution:
I = â« x cos -1 1 - x 2 1 + x 2 dx; (x > 0)
Let, x = tan Î¸ â dx = sec 2 Î¸ d Î¸
cos -1 1 - x 2 1 + x 2 = cos -1 cos 2 Î¸
= 2 Î¸
= 2 tan -1 x
I = â« tan Î¸ 2 Î¸ Â· sec 2 Î¸ d Î¸
= 2 Î¸ â« tan Î¸ sec 2 Î¸ d Î¸ - â« â« tan Î¸ sec 2 Î¸ d Î¸ d d Î¸ Î¸ d Î¸
= 2 Î¸ tan 2 Î¸ 2 - â« tan 2 Î¸ 2 1 d Î¸
= 2 2 Î¸ tan 2 Î¸ - â« sec 2 Î¸ - 1 d Î¸
=Î¸ tan 2 Î¸ - tan Î¸ + Î¸ + c
= x 2 tan -1 x - x + tan -1 x + c
= - x + 1 + x 2 tan -1 x + c

Topic: Mathematics, Indefinite Integration, JEE Main