datavorous/entrance-exam-dataset huggingface dataset:
The value of the integral â«0121+3x+121-x614 dx is

Options:

The correct answer is:
2

Solution:
∫0121+3dx1+x21-x614
∫0121+3dx1+x21-x61+x614
Put 1-x1+x=t⇒-2dx1+x2=dt
I=∫1131+3dt-2t64=-1+32×-2t113=1+3 3-1=2

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral∫014t31+t8+8t41+t7dt is

Options:
A 128
B 512
C 256
D 1024

The correct answer is:
256

Solution:
Note that ddtt41+t8=4t31+t8+8t41+t7
Hence, the required integral is t41+t801=28

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral $\int_{0}^{1} e^{x^{2}} d x$

Options:
A is less than 1
B is greater than 1
C is less than or equal to 1
D lies in the closed interval $[1, e]$

The correct answer is:
lies in the closed interval $[1, e]$

Solution:
We know that, $0 \leq x \leq 1$ $\Rightarrow \quad x^{2} \leq 1$
$\begin{array}{ll}\Rightarrow & e^{x^{2}} \leq e \\ \Rightarrow & \int_{0}^{1} e^{x^{2}} d x \leq \int_{0}^{1} e d x \\ \Rightarrow & \int_{0}^{1} e^{x^{2}} d x \leq e \\ \therefore \quad \int_{0}^{-} e^{x^{2}} \in[1, e]\end{array}$

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral ∫01xcot-11-x2+x4dx is

Options:
A π4-12loge2
B Ï4-loge2
C Ï2-loge2
D Ï2-12loge2

The correct answer is:
π4-12loge2

Solution:
Given integral, I=∫01xcot-11-x2+x4dx

Put x2=t, ⇒2xdx=dt

⇒I=12∫01cot-11-t+t2dt

⇒I=12∫01tan-111-t+t2dt

⇒I=12∫01tan-1t-t-11+tt-1dt

Using tan-1a-tan-1b=tan-1a-b1+a·b, we get

⇒I=12∫01tan-1tdt-∫01tan-1t-1dt

Using ∫abfxdx=∫abfa+b-xdx, we get

⇒I=12∫01tan-1tdt-∫01tan-11-t-1dt

⇒I=12∫01tan-1tdt-∫01tan-1-tdt

Now, using tan-1-a=-tan-1a, we get

⇒I=12∫01tan-1tdt+∫01tan-1tdt

⇒I=122∫01tan-1tdt

⇒I=∫011⋅tan-1tdt

Now applying the integration by parts i.e. ∫abu·vdx=u∫abvdx-∫abddxu∫vdxdx+c

⇒I=tan-1t⋅t|01-∫0111+t2⋅tdt

Put 1+t2=u, ⇒2tdt=du

⇒I=tan-11⋅1-0-12∫121udu

⇒I=π4-12log2u​12

⇒I=π4-12loge2.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The value of the integral ∫01xdx(1+x)(1+3x)(3+x) is:

Options:
A π41−32
B π81−36
C π81−32
D π41-36

The correct answer is:
π81−32

Solution:
Let x=t

⇒x=t2, dx=2tdt

Let I=∫01xdx(1+x)(1+3x)(3+x)

I=∫01 2t2dtt2+13t2+1t2+3

=∫01 3t2+1−t2+1dtt2+13t2+1t2+3

=∫01 1t2+3t2+1−1t2+33t2+1dt

=∫01 dt2t2+1−183dt3t2+1−38dtt2+3

=12tan−1⁡t−338×3tan−1⁡3t−383tan−1⁡t301

=π8−38×π3−38×π6

=π8−3π16

=π81−32

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (27 Aug Shift 2)
The value of the integral ∫0π260sin6xsinxdx is equal to

Options:

The correct answer is:
104

Solution:
Let I=∫0π260sin6xsinxdx

⇒I=∫0π2602sin3xcos3xsinxdx

⇒I=60∫0π223sinx-4sin3x4cos3x-3cosxsinxdx

⇒I=120∫0π23-4sin2x4cos2x-3cosxdx

⇒I=120∫0π23-4sin2x1-4sin2xcosxdx

Now let sinx=t⇒cosxdx=dt and limit changes to 0 to 1,

So, I=120∫013-4t21-4t2dt

⇒I=120∫013-16t2+16t4 dt

⇒I=1203t-16t33+16t5501

⇒I=1203-163+165=120×45-80+4815=8×13=104

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (28 Jul Shift 2)
The value of the integral Î=â«0Ï2cosâ¡x-sinâ¡x10-x2+Ïx2dx is equal to

Options:
A π2
B π
C 0
D 4π

The correct answer is:
0

Solution:
Applying a+b-x, property,
Ι=∫0π2sin⁡x-cos⁡x10+π2-xxdx
⇒Ι=-Ι
⇒2Ι=0⇒Ι=0

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral ∫04x2x2-4x+8dx is equal to

Options:

The correct answer is:
4

Solution:
Let I=∫04x2xx-4+8dx

Using ∫abfxdx=∫abfa+b-xdx, we get,
I=∫044-x24-x-x+8dx
Adding both the equations, we get,

2I=∫042x2-4x+8x2-4x+8dx=8

⇒I=4

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral ∫04x3-6x2+12x-4+x-2cos⁡x-2dx is equal to

Options:
A 12
B 16
C 0
D 10

The correct answer is:
16

Solution:
Let, I=∫04x-23+x-2cos⁡x-2+4dx
Let x-2=t⇒dx=dt
⇒I=∫-22t3+tcos⁡t+4dt
(As t3+tcos⁡t is an odd function)

⇒I=2∫024dt=16

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral ∫0π4xdxsin42x+cos42x equals:

Options:
A 2π28
B 2π216
C 2π232
D 2π264

The correct answer is:
2π232

Solution:
Let, I=∫0π4xdxsin42x+cos42x

Let, 2x=t then dx=12dt

⇒I=14∫0π2tdtsin4t+cos4t

⇒I=14∫0π2π2−tdtsin4π2−t+cos4π2−t

⇒I=14∫0π2π2dtsin4t+cos4t−I

⇒2I=π8∫0π2dtsin4t+cos4t

⇒2I=π8∫0π2sec4tdttan4t+1

Let, tant=y then sec2tdt=dy

⇒2I=π8∫0∞1+y2dy1+y4

⇒I=π16∫0∞1+1y2y2+1y2dy

Putting, y−1y=p

⇒I=π16∫−∞∞dpp2+22

⇒I=π162tan−1p2−∞∞

⇒I=π2162

⇒I=2π232

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2024 (01 Feb Shift 1)
The value of the integral ∫08x2x2-8x+32dx is equal to

Options:

The correct answer is:
8

Solution:
Let, I=∫08x2xx-8+32dx
So, I=∫088-x2-x8-x+32dx∫abfxdx=∫abfa+b-xdx
Adding both equations, we get,

2I=∫082x2-8x+32x2-8x+32dx=16

⇒I=8

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral $\int_0^{0.9}[x-2[x]] d x$, where [.] denotes the greatest integer function is

Options:
A
$0.9$

B
$1.8$

C
$-0.9$

D
0

The correct answer is:
0

Solution:
Since $\int_0^a[x]=0$ where $0 \leq a \leq 1$
$$
\therefore \int_0^{0.9}[x-2[x]] d x=0
$$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2012 (19 May Online)
The value of the integral Î=â«0100Ïdx1+esinâ¡x is equal to

Options:
A 100π
B 50π
C 25π
D 10π

The correct answer is:
50π

Solution:
As the period of fx=11+esinx is 2π
So, Ι=50∫02πdx1+esin⁡x
Applying a+b-x property and adding, we get,
2Ι=50∫02π11+esin⁡x+11+e-sin⁡xdx
(As sin2π-x=-sin⁡x )
=50∫02πdx=100π
⇒Ι=50π

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral Ι=∫0∞dx1+x20201+x2 is equal to kπ, then the value of 16k is equal to

Options:

The correct answer is:
4

Solution:
Put tan-1x=θ
⇒x=tan⁡θ
⇒dx=sec2θdθ
⇒Ι=∫0π/2sec2θ1+tan2020θsec2θdθ
=∫0π2dθ1+tan2020θ
Ι=∫0π2cos2020θcos2020θ+sin2020θdθ
Applying a+b-x property and adding, we get,
2Ι=∫0π2cos2020θ+sin2020θcos2020θ+sin2020θdθ
⇒2Ι=θ0π2=π2-0
⇒Ι=π4
⇒k=14⇒16k=4

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral ∫0πecos⁡xsin⁡x1+ecot⁡xdx is equal to

Options:
A e+1
B 1-e
C e-1
D -1+e2

The correct answer is:
e-1

Solution:
Let, Ι=∫0πecos⁡xsin⁡x1+ecot⁡xdx … 1
Ι=∫0πecos⁡xsin⁡x1+e-cot⁡xdx … 2 Applying a+b-x property
1+2, we get,
2I=∫0πecos⁡xsin⁡xdx
Putting cos⁡x=t and
-sin⁡xdx=dt
I=12∫1-1-etdt
I=12∫-11(et)dt
I=∫01etdt=e-1 As et is even
I=e-1

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral ∫0πsin2xdx is ________.

Options:

The correct answer is:
2

Solution:
Put 2x=t⇒2dx=dt

⇒I=12∫02πsintdt

=∫0πsintdt

=∫0πsint dt

=-cost0π

=2

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (26 Feb Shift 1)
The value of the integral Ι=∫0πsin⁡x+cos⁡xdx, (where . denotes the greatest integer function) is equal to

Options:
A 1
B 2
C π
D 2π

The correct answer is:
π

Solution:
Let, y=sin⁡x+cos⁡x
⇒y2=1+sin2x∈1,2
⇒y∈1,2
⇒sin⁡x+cos⁡x=1
Thus, Ι=∫0π1·dx=π-0
=π

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral ∫-11dx1+x21+ex is equal to

Options:
A π4
B π2
C π
D 0

The correct answer is:
π4

Solution:
Let, I=∫-11dx1+x21+ex

As, ∫-aafxdx=∫0afx+f-xdx
So, I=∫0111+x211+ex+11+e-xdx
⇒I=∫0111+x2=tan-1x01=π4

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral $\int_{-1}^1 \frac{d}{d x}\left(\tan ^{-1} \frac{1}{x}\right) d x$ is

Options:
A $\frac{\pi}{2}$
B $\frac{\pi}{4}$
C $-\frac{\pi}{2}$
D None of these

The correct answer is:
$-\frac{\pi}{2}$

Solution:
$\int_{-1}^1 \frac{d}{d x}\left(\tan ^{-1} \frac{1}{x}\right) d x=-2\left[\tan ^{-1}(x)\right]_0^1=-\frac{\pi}{2}$

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral $\int_{-1}^{1}\left\{\frac{x^{2013}}{e^{|x|}\left(x^{2}+\cos x\right)}+\frac{1}{e^{|x|}}\right\} d x$ is equal to

Options:
A 0
B $1-e^{-1}$
C $2 e^{-1}$
D $2\left(1-e^{-1}\right)$

The correct answer is:
$2\left(1-e^{-1}\right)$

Solution:
Let $\quad I=\int_{-1}^{1}\left\{\frac{x^{2013}}{e^{\mid x\left|(x^{2}+\cos x\right)}}+\frac{1}{e^{|x|}}\right\} d x$
$\Rightarrow \quad I=\int_{-1}^{1} \frac{x^{2013}}{e^{|x|+\left(x^{2}+\cos x\right)}} d x+\int_{-1}^{1} \frac{1}{e^{|x|}} d x$
Here, $\frac{x^{2013}}{e^{|x|}\left(x^{2}+\cos x\right)}$ is an odd function
and $\frac{1}{e^{\mid x|}}$ is an even function. $\left\{\because \int_{-a}^{a} f(x) d x=\left\{\begin{array}{cc}2 \int_{0}^{9}(x) d x ; & f(x) \text { is even } \\ 0, & f(x) \text { is odd }\end{array}\right\}\right.$
$\left.\therefore \quad 1=0+2 \int_{0}^{1} e^{-x} d x=-2 (e^{-x})\right\}_{0}^{1}=-2\left(e^{-1}\right)$
$=2\left(1-e^{-1}\right)$

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral ∫-11loge1-x+1+xdx is equal to:

Options:
A 12loge2+π4-32
B 2loge2+π4-1
C loge2+π2-1
D 2loge2+π2-12

The correct answer is:
loge2+π2-1

Solution:
Let

I=∫-11loge1-x+1+xdx

⇒I=2∫01loge1-x+1+x⏟I·1⏟IIdx

⇒I=2xloge1-x+1+x01-∫01x·11-x+1+x·121+x-121-xdx

⇒I=2loge2-22∫01x1-x-1+x1-x+1+x1-x2dx

⇒I=2loge2-∫01x1-x-1+x2-2x1-x2dx

⇒I=loge2+12∫012-21-x21-x2 dx

⇒I=loge2+∫011-1-x21-x2dx

⇒I=loge2+sin-1x-x01

⇒I=loge2+π2-1

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (20 Jul Shift 1)
The value of the integral ∫-11logx+x2+1dx is:

Options:
A 2
B 0
C -1
D 1

The correct answer is:
0

Solution:
Let I=∫-11logx+x2+1dx

Consider fx=logx+x2+1

Now, f-x=log-x+1+x2

⇒f-x=-logx+1+x2

⇒f-x=-fx

So, fx is an odd function.

∴ I=0

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (25 Jul Shift 2)
The value of the integral ∫1/22tan-1xxdx is equal to

Options:
A π loge2
B 12loge2
C π4loge2
D π2loge2

The correct answer is:
π2loge2

Solution:
Let,

I=∫122tan-1xxdx ........1

Now, let x=1t⇒dx=-1t2dt

So,

I=-∫212tan-11t1t×1t2dt

⇒I=-∫212tan-11t×1tdt

⇒I=∫122cot-1t×1tdt

⇒I=∫122cot-1x×1xdx .......2

Now adding 1 & 2, we get,

2I=∫1221xcot-1x+tan-1xdx

⇒2I=∫1221x×π2dx

⇒2I=π2lnx122

⇒2I=π2loge2-loge12

⇒2I=π22loge2

⇒I=π2loge2

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2023 (29 Jan Shift 2)
The value of the integral $\int_{-1}^2 \log _e\left(x+\sqrt{x^2+1}\right) d x$ is

Options:
A $\sqrt{5}-\sqrt{2}+\log _e\left(\frac{7+4 \sqrt{5}}{1+\sqrt{2}}\right)$
B $\sqrt{5}-\sqrt{2}+\log _e\left(\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)$
C $\sqrt{2}-\sqrt{5}+\log _e\left(\frac{7+4 \sqrt{5}}{1+\sqrt{2}}\right)$
D $\sqrt{2}-\sqrt{5}+\log _e\left(\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)$

The correct answer is:
$\sqrt{2}-\sqrt{5}+\log _e\left(\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)$

Solution:
$\begin{aligned} & I=\int_{-1}^2 \log _e\left(x+\sqrt{x^2+1}\right) d x \\ & =x \log _e\left(x+\sqrt{x^2+1}\right)-\int_{-1}^2\left(\frac{1+\frac{x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}}\right) d x \\ & =x \log _e\left(x+\sqrt{x^2+1}\right)-\int_{-1}^2 \frac{x}{\sqrt{x^2+1}} d x\end{aligned}$
$\begin{aligned} & =x \log _e\left(x+\sqrt{x^2+1}\right)-\left.\sqrt{x^2+1}\right|_{-1} ^2 \\ & =\left(2 \log _e(2+\sqrt{5})-\sqrt{5}\right) \\ & -\left(-\log _e(-1+\sqrt{2})-\sqrt{2}\right) \\ & =\log _e(2+\sqrt{5})^2-\sqrt{5}+\log _e(\sqrt{2}-1)+\sqrt{2} \\ & =\log _e(2+\sqrt{5})^2-\sqrt{5}+\log _e(\sqrt{2}-1)+\sqrt{2}\end{aligned}$
$\begin{aligned} & =\sqrt{2}-\sqrt{5}+\log _e\left(\frac{(2+\sqrt{5})^2}{\sqrt{2+1}}\right) \\ & =\sqrt{2}-\sqrt{5}+\log _e\left(\frac{9+4 \sqrt{5}}{\sqrt{2+1}}\right)\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2024 (09 Apr Shift 2)
The value of the integral, ∫13x2-2x-2dx, where x denotes the greatest integer less than or equal to x, is

Options:
A -4
B -5
C -2-3+1
D -2-3-1

The correct answer is:
-2-3-1

Solution:
∫13x-12-3dx

=∫02x2dx-3∫13dx

=∫010dx+∫121·dx+∫232·dx+∫323·dx-6

=2-1+23-2+32-3-6

=-2-3-1

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The value of the integral $\int_{-2}^{2}(1+2 \sin x) e^{|x|} d x$ is
equal to

Options:
A 0
B $e^{2}-1$
C $2\left(e^{2}-1\right)$
D 1

The correct answer is:
$2\left(e^{2}-1\right)$

Solution:
$\int_{-2}^{2}(1+2 \sin x) e^{|x|} d x$
$=\int_{-2}^{2} e^{|x|} d x+2 \int_{-2}^{2} \sin x e^{|x|} d x$
(Since, first function is even and second function is odd). $=2 \int_{0}^{2} e^{|x|} d x+2 \text{ instead of ,} 0=2\left[e^{|x|}\right]_{0}^{2}=2\left(e^{2}-1\right)$

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral∫-π2π2dx1+exsin6x+cos6x is equal to

Options:
A 2π
B 0
C π
D π2

The correct answer is:
π

Solution:
Let, I=∫-π2π2dx1+exsm6x+cos6x ...i

Now we know ∫abfxdx=∫abfa+b-xdx

I=∫-π2π2exdx1+exsin6x+cos6x ...ii

Adding i and ii

2I=∫-π2π21+exdx1+exsin6x+cos6x

I=12×2∫0π2dxsin2x+cos2xsin4x+cos4x-sin2xcos2x

=∫0π2dxsin2x+cos2x2−3sin2xcos2x=∫0π2dx1−34sin22x

=∫0π24sec22xdx41+tan22x-3tan22x=2∫0π44sec22x4+tan22xdx

Let tan2x=t and 2sec22xdx=dt

I=4∫0∞dt22+t2=42tan−1t20∞ =2π2-0=π

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (24 Jun Shift 2)
The value of the integral ∫-π2π2sin4⁡x1+ln⁡2+sin⁡x2-sin⁡xdx is

Options:
A 34
B 38π
C 0
D 316Ï

The correct answer is:
38π

Solution:
I=∫-π2π2sin4⁡x1+ln⁡2+sin⁡x2-sin⁡xdx......i

I=2∫0π2sin4xdx ............i,∵sin4x is an even function

∵∫-π2π2sin4xln⁡2+sinx2-sinxdx=0, ∵sin4xln⁡2+sinx2-sinx is odd function

Let ∫0π2sin4x dx=m

⇒m=∫0π2sin4π2-xdx=∫0π2cos4xdx

Adding both, we get

⇒2m=∫0π2sin4xdx+∫0π2cos4xdx

⇒2m=∫0π2sin4x+cos4xdx

⇒2m=sin2x+cos2x2-2sin2xcos2x

⇒2m=1-sin22x2

⇒2m=∫0π21-121-cos4x2dx=∫0π234+cos4x4dx

⇒2m=3x4+sin4x160π2=3π8

⇒m=3π16

Substituting in equation i, we get

I=2m=3π8

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2018 (15 Apr)
The value of the integral
$$
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^4 x\left(1+\log \left(\frac{2+\sin x}{2-\sin x}\right)\right) d x \text { is }
$$

Options:
A
$\frac{3}{16} \pi$

B
0

C
$\frac{3}{8} \pi$

D
$\frac{3}{4}$

The correct answer is:
$\frac{3}{8} \pi$

Solution:
Let
$$
\begin{aligned}
I &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^4 x\left(1+\log \left(\frac{2+\sin x}{2-\sin x}\right)\right) d x \ldots .(1) \\
\Rightarrow I &\left.=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^4(-x)\right)\left(1+\log \left(\frac{2+\sin (-x)}{2-\sin (-x)}\right)\right) \cdot d x \\
&=\left[\because \int_a^b f(x) . d x=\int_a^b f(a+b-x) \cdot d x\right]
\end{aligned}
$$
$$
\begin{aligned}
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\sin ^4 x\right)\left(1+\log \left(\frac{2-\sin x}{2+\sin x}\right)\right) \cdot d x \\
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^4 x\left(1-\log \left(\frac{2+\sin x}{2-\sin x}\right)\right) \cdot d x \ldots .(2)
\end{aligned}
$$
After adding equation (1) and (2) we get,
$$
\begin{aligned}
&2 I=2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^4 x \cdot d x \\
&2 I=4 \int_0^{\frac{\pi}{2}} \sin ^4 x \cdot d x \\
&I=2 \int_0^{\frac{\pi}{2}} \sin ^4 x \cdot d x=\frac{\frac{3}{2} \times \frac{1}{2} \times \pi}{2 \times 2}=\frac{3 \pi}{8} \\
&{[\text { By Gamma function] }}
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
The value of the integral $\int_{-2}^{2} \frac{\sin ^{2} x}{\left[\frac{x}{\pi}\right]+\frac{1}{2}} d x$

(where $[x]$ denotes the greatest integer less than or equal to x) is

Options:
A 0
B $\sin 4$
C 4
D $4-\sin 4$

The correct answer is:
0

Solution:
Let $f(x)=\frac{\sin ^{2} x}{\left[\frac{x}{\pi}\right]+\frac{1}{2}}$.

$\begin{array}{l}

\text { So, } f(-x)=\frac{\sin ^{2}(-x)}{\left[\frac{-x}{\pi}\right]+\frac{1}{2}} \quad \because[-x]=-1-[x] \\

\Rightarrow f(-x)=\frac{\sin ^{2} x}{-1-\left[\frac{x}{\pi}\right]+\frac{1}{2}}=\frac{\sin ^{2} x}{-\frac{1}{2}-\left[\frac{x}{\pi}\right]}=-f(x)

\end{array}$

$\Rightarrow f(x)$ is odd function

Hence, $\int_{-2}^{2} f(x) d x=0$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2019 (11 Jan Shift 1)
The value of the integral ∫-22x3+xexx+1dx is equal to

Options:
A 5e2
B 3e-2
C 4
D 6

The correct answer is:
6

Solution:
Given,

fx=x3+xexx+1dx

and ∫-22fxdx=∫02fx+f-xdx

=∫02x3+xexx+1+-x3-xe-x-x+1dx

=∫02x3+xexx+1+x3+xe-xx+1dx

=∫02x3+xex2+1+x3+xe-x2+1dx

I=∫02x3+x1+ex2+ex2x3+x1+ex2dx

=∫02x3+xdx

=x44+x2202

=4+2=6

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (27 Jun Shift 1)
The value of the integral Ι=∫2x9+x10x2+x33dx is equal to (where, C is the constant of integration)

Options:
A x421+x2+C
B x62x+12+C
C x4x+12+C
D x62x+13+C

The correct answer is:
x421+x2+C

Solution:
Taking x12 common from denominator,
Ι=∫2x-3+x-2x-2+x-13dx
Let, x-2+x-1=t
⇒-2x-3-x-2dx=dt
∴Ι=-∫dtt3
=t-22+C
=12x-2+x-12+C
=x421+x2+C

Topic: Mathematics, Indefinite Integration, JEE Main
The value of the integral â«-3Ï3Ïsin3xdx is equal to

Options:
A π
B 8π
C 1
D 8

The correct answer is:
8

Solution:
The period of the function sin3x is π, thus I=6∫0πsin3xdx
⇒I=6∫0π3sin⁡x-sin⁡3x4dx As sin3x=3sin⁡x-4sin3x
⇒I=64∫0π3sin⁡x-sin3xdx
⇒I=32-3cos⁡x0π+cos3x30π
=323--3+-13-13
=326-23=32×163=8

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral, $\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x$ is

Options:
A
1/2

B
3/2

C
2

D
1

The correct answer is:
3/2

Solution:
$I=\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x$
$I=\int_3^6 \frac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{x}} d x$
$2 I=\int_3^6 d x=3 \Rightarrow I=\frac{3}{2}$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2006
The value of the integral ∫-44exxdx is equal to (where . denotes the fractional part function)

Options:
A e4
B e4+1
C e4-1
D e2

The correct answer is:
e4-1

Solution:
Let, I=â«-44exxdx
I=â«04exx+-xdx
I=â«04exdx=e4-1

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral ∫-π4π4x+π42-cos2xdx is :

Options:
A π26
B π2123
C π233
D π263

The correct answer is:
π263

Solution:
Let,

I=∫-π4π4x+π42-cos2xdx …1

Now replacing x→-x we get,

⇒I=∫-π4π4-x+π42-cos2xdx …2

Now both the equation we get,

⇒2I=∫-π4π4π22-cos2xdx

⇒I=π4·2∫0π4dx2-cos2xdx as cos2x is even function

⇒I=π4·2∫0π41+tan2xdx21+tan2x-1-tan2x

⇒I=π4·2∫0π4sec2xdx3tan2x+1

Now let tanx=t⇒sec2xdx=dt

⇒I=π2∫01dt3t2+1

⇒I=π23tan-13

⇒I=π263

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2023 (01 Feb Shift 2)
The value of the integral ∫410x2x2−28x+196+x2dx, where x denotes the greatest integer less than or equal to x, is

Options:
A 13
B 6
C 7
D 3

The correct answer is:
3

Solution:
Let I= ∫410x2x2-28x+196+x2dx

⇒I= ∫410x214-x2+x2 dx ...i

Use ∫abfxdx=∫abfa+b-xdx

I= ∫41014-x2x2+14-x2 dx ...ii

By adding equations i & ii, we get

2I= ∫41014-x2+x2x2+14-x2 dx

⇒2I= ∫410dx

⇒2I=6

⇒I=3

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2016 (10 Apr Online)
The value of the integral ∫ 4 10 [ x 2 ] [ x 2 −28x+196 ]+[ x 2 ] dx, where x denotes the greatest integer less than or equal to x, is

Options:
A 13
B 6
C 7
D 3

The correct answer is:
3

Solution:
Let I= ∫410x2x2-28x+196+x2dx ........(i)
Use ∫abfxdx=∫abfa+b-xdx
I= ∫41014-x2x2+14-x2 dx ......(ii)
(i) + (ii)
we get 2I= ∫41014-x2+x2x2+14-x2 dx
⇒2I= ∫410dx
⇒2I=6
⇒I=3

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral 48π4∫0π3πx22-x3sinx1+cos2xdx is equal to ______.

Options:

The correct answer is:
6

Solution:
Let I=48π4∫0π3π2x2-x3sinx1+cos2xdx …1

Using ∫abfxdx=∫abf(a+b-x)dx

We get I=48π4∫0π3π2π-x2-π-x3sin0+π-x1+cos20+π-xdx

I=48π4∫0π3π32-3π2x+3πx22-π3+x3+3π2x-3πx2sinx1+cos2xdx ..............equation 2

Now adding equation (1) + equation (2)

2I=48π4∫0π3π32-π3sinx1+cos2xdx

2I=48π4×π32∫0πsinx1+cos2xdx

2I=24π×2×∫0π2sinx1+cos2xdx

Let cosx=t⇒-sinxdx=dt

So I=24π×∫10-dt1+t2

⇒I=-24πtan1t10

⇒I=-24π0-π4=6

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (26 Jun Shift 1)
The value of the integral ∫ −a a e x 1+ e x dx is

Options:
A ea2
B a
C e-a2
D a2

The correct answer is:
a

Solution:
∫ −a a e x 1+ e x dx​ =  ∫ 0 a e x 1+ e x + e −x 1+ e −x dx ∵ ∫ −a a f(x) dx=  ∫ 0 a f(x) + f(−x)  dx
=  ∫ 0 a e x 1+ e x + 1 1+ e x  dx
=  ∫ 0 a dx=a

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral $a \sum_{k=1}^n \int_0^1 f(k-1+x) d x$ is

Options:
A $\int_0^1 f(x) d x$
B $\int_0^2 f(x) d x$
C $\int_0^n f(x) d x$
D $n \int_0^{+1} f(x) d x$

The correct answer is:
$\int_0^n f(x) d x$

Solution:
$\begin{aligned} & \text { Let } I=\int_0^1 f(k-1+x) d x \\ & \Rightarrow \quad I=\int_{k-1}^k f(t) d t, \text { where } t=k-1+x \Rightarrow \quad I=\int_{k-1}^k f(x) d x \\ & \therefore \sum_{k=1}^n \int_{k-1}^k f(x) d x=\int_0^1 f(x) d x+\int_1^2 f(x) d x+\ldots .+\int_{n-1}^n f(x) d x \\ & =\int_0^n f(x) d x\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral ∫cosec2x-2019cos2019x dx is equal to (where C is the constant of integration)

Options:
A cot⁡xcos⁡x2019+C
B -cot⁡xcos⁡x2019+C
C cot⁡xcos⁡x2019+C
D -cot⁡xcos⁡x2019+C

The correct answer is:
-cot⁡xcos⁡x2019+C

Solution:
∫cosec2xcos⁡x-2019dx-∫2019cos⁡x2019dx
=-cot⁡xcos⁡x-2019+∫cot⁡x2019cos⁡x-2020sin⁡xdx-∫2019cos⁡x-2019dx
=-cot⁡xcos⁡x2019+C

Topic: Mathematics, Indefinite Integration, JEE Main
The value of the integral Î=â«dx1+sinx, âxâ0,Ï2 is equal to klnâ¡tanÏ4+x8+c, then the value of k2 is equal to (where, c is the constant of integration)

Options:
A 2
B 12
C 2
D 22

The correct answer is:
2

Solution:
As, 1+sin⁡x=sin2x2+cos2x2+2sinx2cosx2,
=sinx2+cosx22
Ι=∫dxsinx2+cosx2
=12∫dxsinπ4+x2
=12∫cosecπ4+x2dx
=122ln⁡tanπ8+x4+c
=2ln⁡tanπ8+x4+c
⇒k=2

Hence, 2k=2

Topic: Mathematics, Indefinite Integration, JEE Main
The value of the integral ∫dx1+xx-x2 is equal to (where, C is the constant of integration)

Options:
A x+11-x+C
B 2x-11+x+C
C 2x-11-x+C
D x+11+x+C

The correct answer is:
2x-11-x+C

Solution:
Let,I=∫dxx1+x1-x
Put, x=sin2θ and
dx=2sin⁡θcos⁡θdθ
⇒I=∫2sin⁡θcos⁡θdθsin⁡θ1+sin⁡θcos⁡θ
=2∫dθ1+sin⁡θ=2∫1-sin⁡θcos2θdθ
=2tan⁡θ-sec⁡θ+C
=2x1-x-11-x+C=2x-11-x+C

Topic: Mathematics, Indefinite Integration, JEE Main
The value of the integral â«e3sin-1x11-x2+e3cos-1xdx is equal to
(where, c is an arbitrary constant)

Options:
A e3sin-1x3+xe3Ï2+c
B esin-1x+eÏ/2+c
C e3sin-1x3+xe3π2+c
D eÏ2+exÏ2+c

The correct answer is:
e3sin-1x3+xe3π2+c

Solution:
∫e3sin-1x1-x2dx+∫e3sin-1x+cos-1xdx
=∫e3tdt+∫e3π2dx Put sin-1⁡x=t
=e3t3+e3π2⋅x+c

Topic: Mathematics, Indefinite Integration, JEE Main
The value of the integral ∫ex2+1x2x2-1x+1dx is equal to (where C is the constant of integration)

Options:
A ex2+1x+C
B x2ex2+1x+C
C xex2+1x+C
D x⋅ex2+C

The correct answer is:
xex2+1x+C

Solution:
Ι=∫ex2+1x⏟Ι⋅1⏟ΙΙdx+∫ex2+1x2x-1x2⋅xdx
Using integration by parts, we get,
Ι=ex2+1x·x-∫ex2+1x2x-1x2⋅xdx+∫ex2+1x2x-1x2⋅xdx
=xex2+1x+C

Topic: Mathematics, Indefinite Integration, JEE Main
The value of the integral Ι=∫exsin⁡x+cos⁡xdx is equal to ex⋅fx+C, C being the constant of integration. Then the maximum value of y=fx2, ∀x∈R is equal to

Options:
A 0
B -1
C 1
D 12

The correct answer is:
1

Solution:
As we know,

∫exfx+f'xdx=ex⋅fx+C
Thus, ∫exsin⁡x+cos⁡xdx=ex⋅sin⁡x+C
i.e. fx=sin⁡x
Hence, fx2=sinx2: which has the maximum value of ‘1’

Topic: Mathematics, Indefinite Integration, JEE Main
The value of the integral $I=\int_0^1 x(1-x)^n d x$ is

Options:
A
$\frac{1}{n+1}+\frac{1}{n+2}$

B
$\frac{1}{n+1}$

C
$\frac{1}{n+2}$

D
$\frac{1}{n+1}-\frac{1}{n+2}$

The correct answer is:
$\frac{1}{n+1}-\frac{1}{n+2}$

Solution:
$I=\int_0^1 x(1-x)^n d x$
$-I=\int_0^1-x(1-x)^n d x=\int_0^1(1-x-1)(1-x)^n d x$
$=\int_0^1(1-x)^{n+1} d x-\int_0^1(1-x)^n d x$
$=\left[\frac{(1-x)^{n+2}}{-(n+2)}\right]_0^1-\left[\frac{(1-x)^{n+1}}{-(n+1)}\right]_0^1=\frac{1}{n+2}-\frac{1}{n+1}$
$I=\frac{1}{n+1}-\frac{1}{n+2}$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2003
The value of the integral I=â«0Ï2sin3â¡x-cos3â¡x1+sin6â¡xcos6â¡xdx is equal to

Options:

The correct answer is:
0

Solution:
Applying a+b-x property,
I=∫0π2cos3⁡x-sin3⁡x1+cos6⁡xsin6⁡x
⇒I=-I⇒2I=0
⇒I=0

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral I=∫0π4sin⁡x+cos⁡xcos⁡x-sin⁡xdx is equal to

(where, . denotes the greatest integer function)

Options:
A 2
B 22
C 1
D 2-1

The correct answer is:
2-1

Solution:
Let, sin⁡x+cos⁡x=t

⇒I=∫12tdt=∫121dt=2-1

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral I=∫0πx1+tan6⁡xdx, x≠π2 is equal to

Options:
A π2
B π24
C π4
D π22

The correct answer is:
π24

Solution:
Applying a+b-x and adding, we get,
2I=∫0πx+π-x1+tan6⁡xdx
⇒2I=π∫0πdx1+tan6⁡x
⇒2I=π∫0π211+tan6⁡x+11+tan6⁡π-xdx
⇒2I=2π∫0π2dx1+tan6⁡x=2π∫0π2cos6⁡xcos6⁡x+sin6⁡xdx
Again, applying a+b-x and adding, we get,

2I=π∫0π2sin6⁡x+cos6⁡xsin6⁡x+cos6⁡xdx
⇒2I=π∫0π21dx=ππ2
⇒I=π24

Topic: Mathematics, Definite Integration, JEE Main