QBManager

Integral Calculus

86751 For $x > 0$ , If $f (x) = \int_{1}^{x} \frac{\log_{e} t}{(1 + t)} d t$ , then $f (e) + f (\frac{1}{e})$ is equal to

1 1

2 -1

3

\frac{1}{2}

4 0

Explanation:

(C) : Given,
$f (x) = \int_{1}^{x} \frac{\log_{e} t}{(1 + t)} d t$
$f (e) = \int_{1}^{e} \frac{\log_{e} t}{1 + t} d t$
$f (\frac{1}{e}) = \int_{1}^{1 / e} \frac{\log_{e} t}{1 + t} d t$
Let, $t = \frac{1}{u}$ and putting in equation (ii), we get-
$f (\frac{1}{e}) = \int_{1}^{e} \frac{\log (\frac{1}{u})}{1 + \frac{1}{u}} \times \frac{- 1}{u^{2}} du$
$= \int_{1}^{e} \frac{\log u}{u (u + 1)} du$
Using change of variable
$f (\frac{1}{e}) = \int_{1}^{e} \frac{(\log t)}{t (t + 1)} dt$
From equation (i) and (iii), we get-
$f (e) + f (\frac{1}{e}) = \int_{1}^{e} \frac{\log t}{1 + t} d t + \int_{1}^{e} \frac{\log t}{t (1 + t)} \cdot d t = \int_{1}^{e} \frac{\log t}{t} d t$
Let,
$\log t = v Then, \frac{1}{t} dt = dv$
$f (e) + f (\frac{1}{e}) = \int_{0}^{1} vdv = {[\frac{v^{2}}{2}]}_{0}^{1} = \frac{1}{2}$

JEE Main-2021-26.02.2021

Integral Calculus

86752 $\int_{0}^{π / 4} \sqrt{1 - \sin 2 x} d x =$

1

1 + 2 \sqrt{2}

2

\sqrt{2} - 1

3

\sqrt{2} + 1

4

2 \sqrt{2} - 1

Explanation:

(C) : Given,
$I = \int_{0}^{\frac{π}{4}} \sqrt{1 - \sin 2 x} dx$
$= \int_{0}^{\frac{π}{4}} \sqrt{\sin^{2} x + \cos^{2} x - 2 \sin x \cdot \cos x} dx$
$= \int_{0}^{\frac{π}{4}} \sqrt{(\cos x - \sin x)^{2}} d x$
$= [\sin x + \cos]_{0}^{π / 4}$
$= (\sin π / 4 + \cos π / 4) - (\sin 0^{\circ} - \cos 0^{\circ})$
$= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - (0 - 1)$
$= \frac{2}{\sqrt{2}} + 1$
$= \sqrt{2} + 1$

MHT CET-2022

Integral Calculus

86753 The value of $\int_{- π}^{π} \frac{\cos^{2} x}{1 + a^{x}} d x, a > 0$ is

1

π

2

2 π

3

\frac{π}{2}

4

a π

Explanation:

(C) : Given,
$I = \int_{- π}^{π} \frac{\cos^{2} x}{1 + a^{x}} d x$
$∵ \int_{a}^{b} f (x) = \int_{a}^{b} f (a + b - x) d x$
$I = \int_{- π}^{π} \frac{\cos^{2} (π - π - x)}{1 + a^{(π - π - x)}} d x$
$I = \int_{- π}^{π} \frac{\cos^{2} (- x)}{1 + a^{- x}} d x \Rightarrow I = \int_{- π}^{π} \frac{\cos^{2} (- x)}{1 + \frac{1}{a^{x}}}$
$I = \int_{- π}^{π} \frac{a^{x} \cos^{2} (- x)}{1 + a^{x}}$
Adding equation (i) and (ii), we get -
$2 I = \int_{- π}^{π} \frac{(1 + a^{x}) \cos^{2} x}{1 + a^{x}} d x$
$= \int_{- π}^{π} \cos^{2} x d x$
Let, $f (x) = \cos^{2} x$
$f (- x) = \cos^{2} (- x) = \cos^{2} x = f (x)$
$∵ f (x)$ is even function
$∴ 2 I = 2 \int_{0}^{π} \cos^{2} x d x$
$I = \int_{0}^{π} \cos^{2} x d x$
$I = 2 \int_{0}^{π / 2} \cos^{2} x d x$
$I = \int_{0}^{π / 2} (\cos 2 x + 1) d x$
$= {(\frac{\sin 2 x}{2} + x)}_{0}^{π / 2}$
$= 0 + \frac{π}{2} - 0 = \frac{π}{2}$

MHT CET-2022

Integral Calculus

86754 If $\int_{0}^{\frac{π}{2}} \frac{\cot x}{\cot x + cosec x} d x = m (π + n)$ , then ( $m \cdot n)$ equal

1 1

2

- \frac{1}{2}

3 -1

4 +1

Explanation:

(C) : Given,
$\int_{0}^{π / 2} \frac{\cot x}{\cot x + cosec x} d x = m (π + n)$
Let, $I = \int_{0}^{π / 2} \frac{\cot x}{\cot x + cosec x} d x$ $= \int_{0}^{π / 2} \frac{\frac{\cos x}{\sin x}}{\frac{\cos x}{\sin x} + \frac{1}{\sin x}} d x$
$= \int_{0}^{π / 2} \frac{\cos x}{\cos x + 1} d x$
$= \int_{0}^{π / 2} \frac{2 \cos^{2} x / 2 - 1}{2 \cos^{2} x / 2 - 1 + 1} dx$
$= \int_{0}^{π / 2} \frac{2 \cos^{2} x / 2 - 1}{2 \cos^{2} x / 2} dx$
$= \int_{0}^{π / 2} 1 . dx - \frac{1}{2} \int_{0}^{π / 2} \sec^{2} x / 2 dx$
$= \int_{0}^{π / 2} 1 \cdot dx - \frac{1}{2} \int_{0}^{π / 2} \sec^{2} x / 2 dx$
$= [x]_{0}^{π / 2} - \frac{1}{2} [2 \tan x / 2]_{0}^{π / 2}$
$= [\frac{π}{2} - 0] - [\tan \frac{π}{4} - \tan 0]$
$= \frac{π}{2} - 1$
$I = \frac{1}{2} (π - 2)$
From equation (i), we get -
$m (π + n) = \frac{1}{2} (π - 2)$
Comparing both side we get -
$m = \frac{1}{2}, n = - 2$
Now, $m . n = \frac{1}{2} \times - 2 = - 1$

MHT CET-2022

Integral Calculus

86751 For $x > 0$ , If $f (x) = \int_{1}^{x} \frac{\log_{e} t}{(1 + t)} d t$ , then $f (e) + f (\frac{1}{e})$ is equal to

1 1

2 -1

3

\frac{1}{2}

4 0

Explanation:

(C) : Given,
$f (x) = \int_{1}^{x} \frac{\log_{e} t}{(1 + t)} d t$
$f (e) = \int_{1}^{e} \frac{\log_{e} t}{1 + t} d t$
$f (\frac{1}{e}) = \int_{1}^{1 / e} \frac{\log_{e} t}{1 + t} d t$
Let, $t = \frac{1}{u}$ and putting in equation (ii), we get-
$f (\frac{1}{e}) = \int_{1}^{e} \frac{\log (\frac{1}{u})}{1 + \frac{1}{u}} \times \frac{- 1}{u^{2}} du$
$= \int_{1}^{e} \frac{\log u}{u (u + 1)} du$
Using change of variable
$f (\frac{1}{e}) = \int_{1}^{e} \frac{(\log t)}{t (t + 1)} dt$
From equation (i) and (iii), we get-
$f (e) + f (\frac{1}{e}) = \int_{1}^{e} \frac{\log t}{1 + t} d t + \int_{1}^{e} \frac{\log t}{t (1 + t)} \cdot d t = \int_{1}^{e} \frac{\log t}{t} d t$
Let,
$\log t = v Then, \frac{1}{t} dt = dv$
$f (e) + f (\frac{1}{e}) = \int_{0}^{1} vdv = {[\frac{v^{2}}{2}]}_{0}^{1} = \frac{1}{2}$

JEE Main-2021-26.02.2021

Integral Calculus

86752 $\int_{0}^{π / 4} \sqrt{1 - \sin 2 x} d x =$

1

1 + 2 \sqrt{2}

2

\sqrt{2} - 1

3

\sqrt{2} + 1

4

2 \sqrt{2} - 1

Explanation:

(C) : Given,
$I = \int_{0}^{\frac{π}{4}} \sqrt{1 - \sin 2 x} dx$
$= \int_{0}^{\frac{π}{4}} \sqrt{\sin^{2} x + \cos^{2} x - 2 \sin x \cdot \cos x} dx$
$= \int_{0}^{\frac{π}{4}} \sqrt{(\cos x - \sin x)^{2}} d x$
$= [\sin x + \cos]_{0}^{π / 4}$
$= (\sin π / 4 + \cos π / 4) - (\sin 0^{\circ} - \cos 0^{\circ})$
$= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - (0 - 1)$
$= \frac{2}{\sqrt{2}} + 1$
$= \sqrt{2} + 1$

MHT CET-2022

Integral Calculus

86753 The value of $\int_{- π}^{π} \frac{\cos^{2} x}{1 + a^{x}} d x, a > 0$ is

1

π

2

2 π

3

\frac{π}{2}

4

a π

Explanation:

(C) : Given,
$I = \int_{- π}^{π} \frac{\cos^{2} x}{1 + a^{x}} d x$
$∵ \int_{a}^{b} f (x) = \int_{a}^{b} f (a + b - x) d x$
$I = \int_{- π}^{π} \frac{\cos^{2} (π - π - x)}{1 + a^{(π - π - x)}} d x$
$I = \int_{- π}^{π} \frac{\cos^{2} (- x)}{1 + a^{- x}} d x \Rightarrow I = \int_{- π}^{π} \frac{\cos^{2} (- x)}{1 + \frac{1}{a^{x}}}$
$I = \int_{- π}^{π} \frac{a^{x} \cos^{2} (- x)}{1 + a^{x}}$
Adding equation (i) and (ii), we get -
$2 I = \int_{- π}^{π} \frac{(1 + a^{x}) \cos^{2} x}{1 + a^{x}} d x$
$= \int_{- π}^{π} \cos^{2} x d x$
Let, $f (x) = \cos^{2} x$
$f (- x) = \cos^{2} (- x) = \cos^{2} x = f (x)$
$∵ f (x)$ is even function
$∴ 2 I = 2 \int_{0}^{π} \cos^{2} x d x$
$I = \int_{0}^{π} \cos^{2} x d x$
$I = 2 \int_{0}^{π / 2} \cos^{2} x d x$
$I = \int_{0}^{π / 2} (\cos 2 x + 1) d x$
$= {(\frac{\sin 2 x}{2} + x)}_{0}^{π / 2}$
$= 0 + \frac{π}{2} - 0 = \frac{π}{2}$

MHT CET-2022

Integral Calculus

86754 If $\int_{0}^{\frac{π}{2}} \frac{\cot x}{\cot x + cosec x} d x = m (π + n)$ , then ( $m \cdot n)$ equal

1 1

2

- \frac{1}{2}

3 -1

4 +1

Explanation:

(C) : Given,
$\int_{0}^{π / 2} \frac{\cot x}{\cot x + cosec x} d x = m (π + n)$
Let, $I = \int_{0}^{π / 2} \frac{\cot x}{\cot x + cosec x} d x$ $= \int_{0}^{π / 2} \frac{\frac{\cos x}{\sin x}}{\frac{\cos x}{\sin x} + \frac{1}{\sin x}} d x$
$= \int_{0}^{π / 2} \frac{\cos x}{\cos x + 1} d x$
$= \int_{0}^{π / 2} \frac{2 \cos^{2} x / 2 - 1}{2 \cos^{2} x / 2 - 1 + 1} dx$
$= \int_{0}^{π / 2} \frac{2 \cos^{2} x / 2 - 1}{2 \cos^{2} x / 2} dx$
$= \int_{0}^{π / 2} 1 . dx - \frac{1}{2} \int_{0}^{π / 2} \sec^{2} x / 2 dx$
$= \int_{0}^{π / 2} 1 \cdot dx - \frac{1}{2} \int_{0}^{π / 2} \sec^{2} x / 2 dx$
$= [x]_{0}^{π / 2} - \frac{1}{2} [2 \tan x / 2]_{0}^{π / 2}$
$= [\frac{π}{2} - 0] - [\tan \frac{π}{4} - \tan 0]$
$= \frac{π}{2} - 1$
$I = \frac{1}{2} (π - 2)$
From equation (i), we get -
$m (π + n) = \frac{1}{2} (π - 2)$
Comparing both side we get -
$m = \frac{1}{2}, n = - 2$
Now, $m . n = \frac{1}{2} \times - 2 = - 1$

MHT CET-2022

Integral Calculus

86751 For $x > 0$ , If $f (x) = \int_{1}^{x} \frac{\log_{e} t}{(1 + t)} d t$ , then $f (e) + f (\frac{1}{e})$ is equal to

1 1

2 -1

3

\frac{1}{2}

4 0

Explanation:

(C) : Given,
$f (x) = \int_{1}^{x} \frac{\log_{e} t}{(1 + t)} d t$
$f (e) = \int_{1}^{e} \frac{\log_{e} t}{1 + t} d t$
$f (\frac{1}{e}) = \int_{1}^{1 / e} \frac{\log_{e} t}{1 + t} d t$
Let, $t = \frac{1}{u}$ and putting in equation (ii), we get-
$f (\frac{1}{e}) = \int_{1}^{e} \frac{\log (\frac{1}{u})}{1 + \frac{1}{u}} \times \frac{- 1}{u^{2}} du$
$= \int_{1}^{e} \frac{\log u}{u (u + 1)} du$
Using change of variable
$f (\frac{1}{e}) = \int_{1}^{e} \frac{(\log t)}{t (t + 1)} dt$
From equation (i) and (iii), we get-
$f (e) + f (\frac{1}{e}) = \int_{1}^{e} \frac{\log t}{1 + t} d t + \int_{1}^{e} \frac{\log t}{t (1 + t)} \cdot d t = \int_{1}^{e} \frac{\log t}{t} d t$
Let,
$\log t = v Then, \frac{1}{t} dt = dv$
$f (e) + f (\frac{1}{e}) = \int_{0}^{1} vdv = {[\frac{v^{2}}{2}]}_{0}^{1} = \frac{1}{2}$

JEE Main-2021-26.02.2021

Integral Calculus

86752 $\int_{0}^{π / 4} \sqrt{1 - \sin 2 x} d x =$

1

1 + 2 \sqrt{2}

2

\sqrt{2} - 1

3

\sqrt{2} + 1

4

2 \sqrt{2} - 1

Explanation:

(C) : Given,
$I = \int_{0}^{\frac{π}{4}} \sqrt{1 - \sin 2 x} dx$
$= \int_{0}^{\frac{π}{4}} \sqrt{\sin^{2} x + \cos^{2} x - 2 \sin x \cdot \cos x} dx$
$= \int_{0}^{\frac{π}{4}} \sqrt{(\cos x - \sin x)^{2}} d x$
$= [\sin x + \cos]_{0}^{π / 4}$
$= (\sin π / 4 + \cos π / 4) - (\sin 0^{\circ} - \cos 0^{\circ})$
$= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - (0 - 1)$
$= \frac{2}{\sqrt{2}} + 1$
$= \sqrt{2} + 1$

MHT CET-2022

Integral Calculus

86753 The value of $\int_{- π}^{π} \frac{\cos^{2} x}{1 + a^{x}} d x, a > 0$ is

1

π

2

2 π

3

\frac{π}{2}

4

a π

Explanation:

(C) : Given,
$I = \int_{- π}^{π} \frac{\cos^{2} x}{1 + a^{x}} d x$
$∵ \int_{a}^{b} f (x) = \int_{a}^{b} f (a + b - x) d x$
$I = \int_{- π}^{π} \frac{\cos^{2} (π - π - x)}{1 + a^{(π - π - x)}} d x$
$I = \int_{- π}^{π} \frac{\cos^{2} (- x)}{1 + a^{- x}} d x \Rightarrow I = \int_{- π}^{π} \frac{\cos^{2} (- x)}{1 + \frac{1}{a^{x}}}$
$I = \int_{- π}^{π} \frac{a^{x} \cos^{2} (- x)}{1 + a^{x}}$
Adding equation (i) and (ii), we get -
$2 I = \int_{- π}^{π} \frac{(1 + a^{x}) \cos^{2} x}{1 + a^{x}} d x$
$= \int_{- π}^{π} \cos^{2} x d x$
Let, $f (x) = \cos^{2} x$
$f (- x) = \cos^{2} (- x) = \cos^{2} x = f (x)$
$∵ f (x)$ is even function
$∴ 2 I = 2 \int_{0}^{π} \cos^{2} x d x$
$I = \int_{0}^{π} \cos^{2} x d x$
$I = 2 \int_{0}^{π / 2} \cos^{2} x d x$
$I = \int_{0}^{π / 2} (\cos 2 x + 1) d x$
$= {(\frac{\sin 2 x}{2} + x)}_{0}^{π / 2}$
$= 0 + \frac{π}{2} - 0 = \frac{π}{2}$

MHT CET-2022

Integral Calculus

86754 If $\int_{0}^{\frac{π}{2}} \frac{\cot x}{\cot x + cosec x} d x = m (π + n)$ , then ( $m \cdot n)$ equal

1 1

2

- \frac{1}{2}

3 -1

4 +1

Explanation:

(C) : Given,
$\int_{0}^{π / 2} \frac{\cot x}{\cot x + cosec x} d x = m (π + n)$
Let, $I = \int_{0}^{π / 2} \frac{\cot x}{\cot x + cosec x} d x$ $= \int_{0}^{π / 2} \frac{\frac{\cos x}{\sin x}}{\frac{\cos x}{\sin x} + \frac{1}{\sin x}} d x$
$= \int_{0}^{π / 2} \frac{\cos x}{\cos x + 1} d x$
$= \int_{0}^{π / 2} \frac{2 \cos^{2} x / 2 - 1}{2 \cos^{2} x / 2 - 1 + 1} dx$
$= \int_{0}^{π / 2} \frac{2 \cos^{2} x / 2 - 1}{2 \cos^{2} x / 2} dx$
$= \int_{0}^{π / 2} 1 . dx - \frac{1}{2} \int_{0}^{π / 2} \sec^{2} x / 2 dx$
$= \int_{0}^{π / 2} 1 \cdot dx - \frac{1}{2} \int_{0}^{π / 2} \sec^{2} x / 2 dx$
$= [x]_{0}^{π / 2} - \frac{1}{2} [2 \tan x / 2]_{0}^{π / 2}$
$= [\frac{π}{2} - 0] - [\tan \frac{π}{4} - \tan 0]$
$= \frac{π}{2} - 1$
$I = \frac{1}{2} (π - 2)$
From equation (i), we get -
$m (π + n) = \frac{1}{2} (π - 2)$
Comparing both side we get -
$m = \frac{1}{2}, n = - 2$
Now, $m . n = \frac{1}{2} \times - 2 = - 1$

MHT CET-2022

Integral Calculus

86751 For $x > 0$ , If $f (x) = \int_{1}^{x} \frac{\log_{e} t}{(1 + t)} d t$ , then $f (e) + f (\frac{1}{e})$ is equal to

1 1

2 -1

3

\frac{1}{2}

4 0

Explanation:

(C) : Given,
$f (x) = \int_{1}^{x} \frac{\log_{e} t}{(1 + t)} d t$
$f (e) = \int_{1}^{e} \frac{\log_{e} t}{1 + t} d t$
$f (\frac{1}{e}) = \int_{1}^{1 / e} \frac{\log_{e} t}{1 + t} d t$
Let, $t = \frac{1}{u}$ and putting in equation (ii), we get-
$f (\frac{1}{e}) = \int_{1}^{e} \frac{\log (\frac{1}{u})}{1 + \frac{1}{u}} \times \frac{- 1}{u^{2}} du$
$= \int_{1}^{e} \frac{\log u}{u (u + 1)} du$
Using change of variable
$f (\frac{1}{e}) = \int_{1}^{e} \frac{(\log t)}{t (t + 1)} dt$
From equation (i) and (iii), we get-
$f (e) + f (\frac{1}{e}) = \int_{1}^{e} \frac{\log t}{1 + t} d t + \int_{1}^{e} \frac{\log t}{t (1 + t)} \cdot d t = \int_{1}^{e} \frac{\log t}{t} d t$
Let,
$\log t = v Then, \frac{1}{t} dt = dv$
$f (e) + f (\frac{1}{e}) = \int_{0}^{1} vdv = {[\frac{v^{2}}{2}]}_{0}^{1} = \frac{1}{2}$

JEE Main-2021-26.02.2021

Integral Calculus

86752 $\int_{0}^{π / 4} \sqrt{1 - \sin 2 x} d x =$

1

1 + 2 \sqrt{2}

2

\sqrt{2} - 1

3

\sqrt{2} + 1

4

2 \sqrt{2} - 1

Explanation:

(C) : Given,
$I = \int_{0}^{\frac{π}{4}} \sqrt{1 - \sin 2 x} dx$
$= \int_{0}^{\frac{π}{4}} \sqrt{\sin^{2} x + \cos^{2} x - 2 \sin x \cdot \cos x} dx$
$= \int_{0}^{\frac{π}{4}} \sqrt{(\cos x - \sin x)^{2}} d x$
$= [\sin x + \cos]_{0}^{π / 4}$
$= (\sin π / 4 + \cos π / 4) - (\sin 0^{\circ} - \cos 0^{\circ})$
$= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - (0 - 1)$
$= \frac{2}{\sqrt{2}} + 1$
$= \sqrt{2} + 1$

MHT CET-2022

Integral Calculus

86753 The value of $\int_{- π}^{π} \frac{\cos^{2} x}{1 + a^{x}} d x, a > 0$ is

1

π

2

2 π

3

\frac{π}{2}

4

a π

Explanation:

(C) : Given,
$I = \int_{- π}^{π} \frac{\cos^{2} x}{1 + a^{x}} d x$
$∵ \int_{a}^{b} f (x) = \int_{a}^{b} f (a + b - x) d x$
$I = \int_{- π}^{π} \frac{\cos^{2} (π - π - x)}{1 + a^{(π - π - x)}} d x$
$I = \int_{- π}^{π} \frac{\cos^{2} (- x)}{1 + a^{- x}} d x \Rightarrow I = \int_{- π}^{π} \frac{\cos^{2} (- x)}{1 + \frac{1}{a^{x}}}$
$I = \int_{- π}^{π} \frac{a^{x} \cos^{2} (- x)}{1 + a^{x}}$
Adding equation (i) and (ii), we get -
$2 I = \int_{- π}^{π} \frac{(1 + a^{x}) \cos^{2} x}{1 + a^{x}} d x$
$= \int_{- π}^{π} \cos^{2} x d x$
Let, $f (x) = \cos^{2} x$
$f (- x) = \cos^{2} (- x) = \cos^{2} x = f (x)$
$∵ f (x)$ is even function
$∴ 2 I = 2 \int_{0}^{π} \cos^{2} x d x$
$I = \int_{0}^{π} \cos^{2} x d x$
$I = 2 \int_{0}^{π / 2} \cos^{2} x d x$
$I = \int_{0}^{π / 2} (\cos 2 x + 1) d x$
$= {(\frac{\sin 2 x}{2} + x)}_{0}^{π / 2}$
$= 0 + \frac{π}{2} - 0 = \frac{π}{2}$

MHT CET-2022

Integral Calculus

86754 If $\int_{0}^{\frac{π}{2}} \frac{\cot x}{\cot x + cosec x} d x = m (π + n)$ , then ( $m \cdot n)$ equal

1 1

2

- \frac{1}{2}

3 -1

4 +1

Explanation:

(C) : Given,
$\int_{0}^{π / 2} \frac{\cot x}{\cot x + cosec x} d x = m (π + n)$
Let, $I = \int_{0}^{π / 2} \frac{\cot x}{\cot x + cosec x} d x$ $= \int_{0}^{π / 2} \frac{\frac{\cos x}{\sin x}}{\frac{\cos x}{\sin x} + \frac{1}{\sin x}} d x$
$= \int_{0}^{π / 2} \frac{\cos x}{\cos x + 1} d x$
$= \int_{0}^{π / 2} \frac{2 \cos^{2} x / 2 - 1}{2 \cos^{2} x / 2 - 1 + 1} dx$
$= \int_{0}^{π / 2} \frac{2 \cos^{2} x / 2 - 1}{2 \cos^{2} x / 2} dx$
$= \int_{0}^{π / 2} 1 . dx - \frac{1}{2} \int_{0}^{π / 2} \sec^{2} x / 2 dx$
$= \int_{0}^{π / 2} 1 \cdot dx - \frac{1}{2} \int_{0}^{π / 2} \sec^{2} x / 2 dx$
$= [x]_{0}^{π / 2} - \frac{1}{2} [2 \tan x / 2]_{0}^{π / 2}$
$= [\frac{π}{2} - 0] - [\tan \frac{π}{4} - \tan 0]$
$= \frac{π}{2} - 1$
$I = \frac{1}{2} (π - 2)$
From equation (i), we get -
$m (π + n) = \frac{1}{2} (π - 2)$
Comparing both side we get -
$m = \frac{1}{2}, n = - 2$
Now, $m . n = \frac{1}{2} \times - 2 = - 1$

MHT CET-2022

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