Test Index

Definite Integrals Part 2

Section: Mathematics

© examsnet.com

Question : 63 of 100

Marks: +1, -0

The value of

\int\limits_{-\pi/2}^{\pi/2} \frac{1+\sin^{2}x}{1+\pi^{\sin x}} \, dx

[26 Aug 2021 Shift 2]

$\frac{\pi}{2}$
$\frac{5\pi}{2}$
$\frac{3\pi}{4}$
$\frac{3\pi}{2}$

Solution:

I=\int\limits_{-\pi/2}^{\pi/2} \frac{1+\sin^{2}x}{1+\pi^{\sin x}} \, d x

...(i)

I=\int\limits_{-\pi/2}^{\pi/2} \frac{1+\sin^{2}(-x)}{1+\pi^{\sin(-x)}}

\left[\because \int\limits_{a}^{b} f(x) \, d x = \int\limits_{a}^{b} f(a+b-x) \, d x\right]

\int\limits_{-\pi/2}^{\pi/2} \frac{1+\sin x}{1+\pi^{-\sin x}} \, d x

I=\int\limits_{-\pi/2}^{\pi/2} \frac{\pi^{\sin x}(1+\sin^{2}x)}{1+\pi^{\sin x}} \, d x

...(ii)

Adding Eqs. (i) and (ii), we get

2I=\int\limits_{-\pi/2}^{\pi/2} \frac{(1+\pi^{\sin x})(1+\sin^{2}x)}{1+\pi^{\sin x}} \, d x

\Rightarrow 2I = \int\limits_{-\pi/2}^{\pi/2} (1+\sin^{2}x) \, d x

\Rightarrow 2I = [x]_{-\pi/2}^{\pi/2} + 2\int\limits_{0}^{\pi/2} \sin^{2}x \, d x

[\because \sin^{2}x

is an even function, so

\int\limits_{-\pi/2}^{\pi/2} \sin^{2} \, d x = 2\int\limits_{0}^{\pi/2} \sin^{2}x \, d x

]

\Rightarrow I = \frac{1}{2} \left( \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) \right) + \frac{1}{2} \int\limits_{0}^{\pi/2} (1-\cos 2x) \, d x

\Rightarrow I = \frac{\pi}{2} + \frac{1}{2} \left[ x - \frac{\sin 2x}{2} \right]_{0}^{\pi/2}

\frac{\pi}{2} + \frac{1}{2} \left[ \left( \frac{\pi}{2} - 0 \right) - (0-0) \right]

I = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}

© examsnet.com

Go to Question:

Prev Question

Next Question