The value of the integral $\int\limits_0^{\pi / 4} \frac{x \mathrm{~d} x}{\sin ^4(2 x)+\cos ^4(2 x)}$ equals :

Take $I=\int\limits_0^{\pi / 4} \frac{x d x}{\sin ^4(2 x)+\cos ^4(2 x)}$ Let $2 x=t$ $2 d x=d t$ $d x=\frac{d t}{2}$ $\begin{aligned} & I=\int\limits_0^{\pi / 2} \frac{t / 2 \cdot 1 / 2 d t}{\sin ^4 t+\cos ^4 t} \\\\ & I=\frac{1}{4} \int\limits_0^{\pi / 2} \frac{t d t}{\sin ^4 t+\cos ^4 t} \\\\ & =\frac{1}{4} \int\limits_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-t\right) d t}{\sin ^4(\pi / 2-t)+\cos ^4(\pi / 2-t)} \\\\ & =\frac{1}{4} \int\limits_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-t\right)}{\sin ^

Definite Integration — Page 16 | JEE Mathematics

Q151

Let

f

be a continuous function satisfying

\int\limits_{0}^{t^{2}}\left(f(x)+x^{2}\right) d x=\dfrac{4}{3} t^{3}, \forall t > 0

. Then

f\left(\dfrac{\pi^{2}}{4}\right)

is equal to :

A

-\pi\left(1+\dfrac{\pi^{3}}{16}\right)

B

\pi\left(1-\dfrac{\pi^{3}}{16}\right)

C

-\pi^{2}\left(1+\dfrac{\pi^{2}}{16}\right)

D

\pi^{2}\left(1-\dfrac{\pi^{2}}{16}\right)

Correct Answer

Option B

Solution

Given that

\int\limits_0^{t^2}\left(f(x)+x^2\right) d x=\frac{4}{3} t^3, \forall t>0

On differentiating using Leibnitz rule, we get

\begin{aligned} & \left(f\left(t^2\right)+t^4\right) \times 2 t=\frac{4}{3} \times 3 t^2 \\\\ & \Rightarrow f\left(t^2\right)+t^4=2 t \\\\ & \Rightarrow f\left(t^2\right)=2 t-t^4 \end{aligned}

On substituting $\dfrac{\pi}{2}$ for $t$ , we get

f\left(\frac{\pi^2}{4}\right)=2\left(\frac{\pi}{2}\right)-\frac{\pi^4}{16}=\pi\left(1-\frac{\pi^3}{16}\right)

Q152

Let

5 f(x)+4 f\left(\dfrac{1}{x}\right)=\dfrac{1}{x}+3, x > 0

. Then

18 \int\limits_{1}^{2} f(x) d x

is equal to :

A

10 \log _{\mathrm{e}} 2+6

B

5 \log _{e} 2-3

C

10 \log _{\mathrm{e}} 2-6

D

5 \log _{\mathrm{e}} 2+3

Correct Answer

Option C

Solution

We have,

5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x>0

..........(i) On replacing $x$ by $\dfrac{1}{x}$ in (i), we get

5 f\left(\frac{1}{x}\right)+4 f(x)=x+3

..........(ii) Now, using Eq. (i) $\times 5-$ (ii) $\times 4$ , we get

\begin{aligned} & 25 f(x)-16 f(x) =\left(\frac{5}{x}+15\right)-(4 x+12) \\\\ &\Rightarrow 9 f(x) =\frac{5}{x}-4 x+3 \\\\ &\Rightarrow f(x) =\frac{1}{9}\left(\frac{5}{x}-4 x+3\right) ...........(iii) \end{aligned}

\begin{aligned} \therefore \quad & 18 \int_1^2 f(x) d x=18 \int_1^2 \frac{1}{9}\left(\frac{5}{x}-4 x+3\right) d x \text{ [Using Eq. (iii)] }\\\\ = & 2 \int_1^2\left(\frac{5}{x}-4 x+3\right) d x \end{aligned}

\begin{aligned} & =2\left[5 \log _e x-4\left(\frac{x^2}{2}\right)+3 x\right]_1^2 \\\\ & =2\left[\left(5 \log _e 2-2(2)^2+3(2)\right)-\left(5 \log 1-2(1)^2+3(1)\right)\right] \\\\ & =2\left[5 \log _e 2-8+6+2-3\right] [\because \log 1=0]\\\\ & =10 \log _e 2-6 \end{aligned}

Q153

Let

f(x)

be a function satisfying

f(x)+f(\pi-x)=\pi^{2}, \forall x \in \mathbb{R}

. Then

\int\limits_{0}^{\pi} f(x) \sin x d x

is equal to :

A

\pi^{2}

B

\dfrac{\pi^{2}}{2}

C

2 \pi^{2}

D

\dfrac{\pi^{2}}{4}

Correct Answer

Option A

Solution

Let $I=\int\limits_0^\pi f(x) \sin x d x$ ..........(i)

\begin{aligned} & =\int\limits_0^\pi f(\pi-x) \sin (\pi-x) d x \\\\ & =\int\limits_0^\pi f(\pi-x) \sin x d x ........(ii) \end{aligned}

On adding Equations (i) and (ii), we get

\begin{aligned} & 2 I=\int\limits_0^\pi[f(x)+f(\pi-x)] \sin x d x \\\\ & \Rightarrow 2 I=\int\limits_0^\pi \pi^2 \sin x d x=\pi^2 \int\limits_0^\pi \sin x d x \\\\ & =\pi^2(-\cos x)_0^\pi=-\pi^2(-1-1)=2 \pi^2 \\\\ & \Rightarrow I=\pi^2 \end{aligned}

Q154

If

\int\limits_0^{\dfrac{\pi}{3}} \cos ^4 x \mathrm{~d} x=\mathrm{a} \pi+\mathrm{b} \sqrt{3}

, where

\mathrm{a}

and

\mathrm{b}

are rational numbers, then

9 \mathrm{a}+8 \mathrm{b}

is equal to :

A 2

B 1

C 3

D

\dfrac{3}{2}

Correct Answer

Option A

Solution

To solve the given integral $\int\limits_0^{\dfrac{\pi}{3}} \cos ^4 x \mathrm{~d} x$ , we'll apply a known power-reduction formula that allows us to express even powers of sine and cosine functions in terms of cosine of multiple angles.

Specifically for $\cos^4 x$ , we can write it in terms of double angles as:

\cos^4 x = \left(\frac{1 + \cos(2x)}{2}\right)^2

We can then expand and simplify the integral using this formula. Let's proceed with this:

\int\limits_0^{\frac{\pi}{3}} \cos^4 x \mathrm{~d}x = \int\limits_0^{\frac{\pi}{3}} \left(\frac{1 + \cos(2x)}{2}\right)^2 \mathrm{~d}x

Now, let's expand the integrand and then integrate term by term:

= \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{4} \cos^2(2x)\right)\mathrm{~d}x

For the $\cos^2(2x)$ term, we again use the power reduction formula:

\cos^2(2x) = \frac{1 + \cos(4x)}{2}

Let's substitute this into the integral and continue:

= \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{4} \left(\frac{1 + \cos(4x)}{2}\right)\right)\mathrm{~d}x

Simplify and integrate:

= \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{8} + \frac{1}{8} \cos(4x)\right)\mathrm{~d}x

= \int\limits_0^{\frac{\pi}{3}} \left(\frac{3}{8} + \frac{1}{2} \cos(2x) + \frac{1}{8} \cos(4x)\right)\mathrm{~d}x

= \left. \left(\frac{3}{8} x + \frac{1}{4} \sin(2x) + \frac{1}{32} \sin(4x)\right) \right|_0^{\frac{\pi}{3}}

Evaluating this from $0$ to $\dfrac{\pi}{3}$ :

= \left(\frac{3}{8} \cdot \frac{\pi}{3} + \frac{1}{4} \sin\left(2 \cdot \frac{\pi}{3}\right) + \frac{1}{32} \sin\left(4 \cdot \frac{\pi}{3}\right)\right) - \left(\frac{3}{8} \cdot 0 + \frac{1}{4} \sin(0) + \frac{1}{32} \sin(0)\right)

= \frac{\pi}{8} + \frac{1}{4} \sin\left(\frac{2\pi}{3}\right) + \frac{1}{32} \sin\left(\frac{4\pi}{3}\right)

$\sin\left(\dfrac{2\pi}{3}\right)$ is $\dfrac{\sqrt{3}}{2}$ and $\sin\left(\dfrac{4\pi}{3}\right)$ is $-\dfrac{\sqrt{3}}{2}$ :

= \frac{\pi}{8} + \frac{1}{4} \cdot \frac{\sqrt{3}}{2} - \frac{1}{32} \cdot \frac{\sqrt{3}}{2}

= \frac{\pi}{8} + \frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{64}

Now, combining terms we get the final result:

= \frac{\pi}{8} + \frac{8\sqrt{3} - \sqrt{3}}{64}

= \frac{\pi}{8} + \frac{7\sqrt{3}}{64}

Now, let's match this result to the form $\mathrm{a}\pi + \mathrm{b}\sqrt{3}$ and find $a$ and $b$ :

a = \frac{1}{8}, \quad b = \frac{7}{64}

Now we find $9a + 8b$ :

9a + 8b = 9 \cdot \frac{1}{8} + 8 \cdot \frac{7}{64}

= \frac{9}{8} + \frac{7}{8}

= \frac{16}{8}

= 2

Therefore, the value of $9a + 8b$ is 2, which correspond to Option A.

Q155

The value of

\int\limits_0^1\left(2 x^3-3 x^2-x+1\right)^{\dfrac{1}{3}} \mathrm{~d} x

is equal to :

A -1

B 2

C 0

D 1

Correct Answer

Option C

Solution

$\begin{aligned} & I=\int_0^1\left(2 x^3-3 x^2-x+1\right)^{1 / 3} d x \\\\ & I=\int_0^1\left((2 x-1)\left(x^2-x-1\right)\right)^{1 / 3} d x \\\\ & I=\int_0^1\left[(2(1-x)-1)\left((1-x)^2-(1-x)-1\right)\right]^{1 / 3} d x \\\\ & I=\int_0^1\left((1-2 x)\left(x^2-x-1\right)\right)^{1 / 3} d x\end{aligned}$ $\begin{aligned} & I=-\int_0^1\left((2 x-1)\left(x^2-x-1\right)\right)^{1 / 3} d x \\\\ & I=-I \\\\ & 2 I=0 \\\\ & I=0\end{aligned}$

Q156

The value of the integral

\int\limits_0^{\pi / 4} \dfrac{x \mathrm{~d} x}{\sin ^4(2 x)+\cos ^4(2 x)}

equals :

A

\dfrac{\sqrt{2} \pi^2}{8}

B

\dfrac{\sqrt{2} \pi^2}{16}

C

\dfrac{\sqrt{2} \pi^2}{32}

D

\dfrac{\sqrt{2} \pi^2}{64}

Correct Answer

Option C

Solution

Take $I=\int\limits_0^{\pi / 4} \dfrac{x d x}{\sin ^4(2 x)+\cos ^4(2 x)}$ Let $2 x=t$ $2 d x=d t$ $d x=\dfrac{d t}{2}$ $\begin{aligned} & I=\int\limits_0^{\pi / 2} \dfrac{t / 2 \cdot 1 / 2 d t}{\sin ^4 t+\cos ^4 t} \\\\ & I=\dfrac{1}{4} \int\limits_0^{\pi / 2} \dfrac{t d t}{\sin ^4 t+\cos ^4 t} \\\\ & =\dfrac{1}{4} \int\limits_0^{\pi / 2} \dfrac{\left(\dfrac{\pi}{2}-t\right) d t}{\sin ^4(\pi / 2-t)+\cos ^4(\pi / 2-t)} \\\\ & =\dfrac{1}{4} \int\limits_0^{\pi / 2} \dfrac{\left(\dfrac{\pi}{2}-t\right)}{\sin ^4 t+\cos ^4 t}\end{aligned}$ $\begin{aligned} & 2 I=\dfrac{1}{4} \int\limits_0^{\pi / 2} \dfrac{\dfrac{\pi}{2}}{\sin ^4 t+\cos ^4 t} d t \\\\ & 2 I=\dfrac{\pi}{8} \int\limits_0^{\pi / 2} \dfrac{d t}{\sin ^4 t+\cos ^4 t} \\\\ & 2 I=\dfrac{\pi}{8} \int\limits_0^{\pi / 2} \dfrac{\sec ^4 t}{1+\tan ^4 t} d t\end{aligned}$ $\begin{aligned} & \text{ Put tant }=y \\\\ & \sec ^2 t d t=d y \\\\ & 2 I=\dfrac{\pi}{8} \int\limits_0^{\infty} \dfrac{\left(1+y^2\right) d y}{1+y^4} \\\\ & =\dfrac{\pi}{8} \int\limits_0^{\infty} \dfrac{1+\dfrac{1}{y^2}}{y^2+\dfrac{1}{y^2}-2+2} d y \\\\ & = \dfrac{\pi}{8} \int\limits_0^{\infty} \dfrac{\left(1+\dfrac{1}{y^2}\right) d y}{2+\left(y-\dfrac{1}{y}\right)^2}\end{aligned}$ $\begin{aligned} & \text{ Put, } y-\dfrac{1}{y}=4 \\\\ & 2 I=\dfrac{\pi}{8} \int\limits_{-\infty}^{\infty} \dfrac{d u}{2+u^2} \\\\ & =\dfrac{\pi}{8 \sqrt{2}}\left[\tan ^{-1} \dfrac{y}{\sqrt{2}}\right]_{-\infty}^{\infty} \\\\ & =\dfrac{\sqrt{2} \pi^2}{32}\end{aligned}$

Q157

If

\int\limits_0^1 \dfrac{1}{\sqrt{3+x}+\sqrt{1+x}} \mathrm{~d} x=\mathrm{a}+\mathrm{b} \sqrt{2}+\mathrm{c} \sqrt{3}

, where

\mathrm{a}, \mathrm{b}, \mathrm{c}

are rational numbers, then

2 \mathrm{a}+3 \mathrm{~b}-4 \mathrm{c}

is equal to :

A 10

B 7

C 4

D 8

Correct Answer

Option D

Solution

\begin{aligned} & \int\limits_0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}} d x=\int\limits_0^1 \frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)} d x \\ & \frac{1}{2}\left[\int\limits_0^1 \sqrt{3+x} d x-\int\limits_0^1(\sqrt{1+x}) d x\right] \end{aligned}

\begin{aligned} & \frac{1}{2}\left[2 \frac{(3+x)^{\frac{3}{2}}}{3}-\frac{2(1+x)^{\frac{3}{2}}}{3}\right]_0^1 \\ & \frac{1}{2}\left[\frac{2}{3}(8-3 \sqrt{3})-\frac{2}{3}\left(2^{\frac{3}{2}}-1\right)\right] \\ & \frac{1}{3}[8-3 \sqrt{3}-2 \sqrt{2}+1] \\ & =3-\sqrt{3}-\frac{2}{3} \sqrt{2}=a+b \sqrt{2}+c \sqrt{3} \\ & a=3, b=-\frac{2}{3}, c=-1 \\ & 2 a+3 b-4 c=6-2+4=8 \end{aligned}

Q158

If

(a, b)

be the orthocentre of the triangle whose vertices are

(1,2),(2,3)

and

(3,1)

, and

\mathrm{I}_1=\int\limits_{\mathrm{a}}^{\mathrm{b}} x \sin \left(4 x-x^2\right) \mathrm{d} x, \mathrm{I}_2=\int\limits_{\mathrm{a}}^{\mathrm{b}} \sin \left(4 x-x^2\right) \mathrm{d} x

, then

36 \dfrac{\mathrm{I}_1}{\mathrm{I}_2}

is equal to :

A 80

B 72

C 66

D 88

Correct Answer

Option B

Solution

Equation of CE

\begin{aligned} & y-1=-(x-3) \\ & x+y=4 \end{aligned}

orthocentre lies on the line

x+y=4

so,

a+b=4

I_1=\int\limits_ a^b x \sin (x(4-x)) d x\quad

..... (i) Using king rule

I_1=\int\limits_ a^b(4-x) \sin (x(4-x)) d x\quad

.... (ii)

\begin{aligned} & \text{ (i) }+ \text{ (ii) } \\ & 2 \mathrm{I}_1=\int\limits_{\mathrm{a}}^{\mathrm{b}} 4 \sin (\mathrm{x}(4-\mathrm{x})) \mathrm{dx} \\ & 2 \mathrm{I}_1=4 \mathrm{I}_2 \\ & \mathrm{I}_1=2 \mathrm{I}_2 \\ & \frac{\mathrm{I}_1}{\mathrm{I}_2}=2 \\ & \frac{36 \mathrm{I}_1}{\mathrm{I}_2}=72 \end{aligned}

Q159

For $$0

A

\dfrac{\pi^2}{\pi+a^2}

B

\dfrac{\pi^2}{\pi-a^2}

C

\dfrac{\pi}{1-\mathrm{a}^2}

D

\dfrac{\pi}{1+\mathrm{a}^2}

Correct Answer

Option C

Solution

\begin{aligned} & I=\int\limits_0^\pi \frac{d x}{1-2 a \cos x+a^2} ; 0

\begin{aligned} & \Rightarrow \mathrm{I}=\int_\limits0^{\pi / 2} \frac{\frac{2 \cdot \sec ^2 \mathrm{x}}{1+\mathrm{a}^2} \cdot \mathrm{dx}}{\tan ^2 \mathrm{x}+\left(\frac{1-\mathrm{a}^2}{1+\mathrm{a}^2}\right)^2} \\ & \Rightarrow \mathrm{I}=\frac{2}{\left(1-\mathrm{a}^2\right)}\left[\frac{\pi}{2}-0\right] \\ & \mathrm{I}=\frac{\pi}{1-\mathrm{a}^2} \end{aligned}$$

Q160

Let

\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}

be defined as

f(x)=a e^{2 x}+b e^x+c x

. If

f(0)=-1, f^{\prime}\left(\log _e 2\right)=21

and

\int_0^{\log _e 4}(f(x)-c x) d x=\dfrac{39}{2}

, then the value of

|a+b+c|

equals

A 16

B 12

C 8

D 10

Correct Answer

Option C

Solution

\begin{array}{ll} \mathrm{f}(\mathrm{x})=a \mathrm{e}^{2 \mathrm{x}}+b \mathrm{e}^{\mathrm{x}}+\mathrm{cx} & \mathrm{f}(0)=-1 \\\\ & \mathrm{a}+\mathrm{b}=-1 \\\\ \mathrm{f}^{\prime}(\mathrm{x})=2 a \mathrm{e}^{2 \mathrm{x}}+b \mathrm{e}^{\mathrm{x}}+\mathrm{c} & \mathrm{f}^{\prime}(\ln 2)=21 \\\\ & 8 \mathrm{a}+2 \mathrm{~b}+\mathrm{c}=21 \\\\ \int\limits_0^{\ln 4}\left(a \mathrm{e}^{2 \mathrm{x}}+b \mathrm{e}^{\mathrm{x}}\right) \mathrm{dx}=\frac{39}{2} & \end{array}

\begin{aligned} & {\left[\frac{\mathrm{ae}^{2 \mathrm{x}}}{2}+\mathrm{be}^{\mathrm{x}}\right]_0^{\ln 4}=\frac{39}{2} \Rightarrow 8 \mathrm{a}+4 \mathrm{~b}-\frac{\mathrm{a}}{2}-\mathrm{b}=\frac{39}{2}} \\\\ & 15 a+6 b=39 \\\\ & 15 a-6 a-6=39 \\\\ & 9 \mathrm{a}=45 \Rightarrow \mathrm{a}=5 \\\\ & b=-6 \\\\ & c=21-40+12=-7 \\\\ & \mathrm{a}+\mathrm{b}+\mathrm{c}-8 \\\\ & |\mathrm{a}+\mathrm{b}+\mathrm{c}|=8 \\\\ \end{aligned}