The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{1+\sin ^{2} x}{1+\pi^{\sin x}}\right) d x$ is:

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Definite Integrals

The value of

\int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{1 + s i n ^{2} x}{1 + π ^{s i n x}}) d x

is:

Answer: (A)

Write

I = \int_{- π /2}^{π /2} \frac{1 + sin ^{2} x}{1 + π ^{s i n x}} d x

. Replacing

x

- x

over this symmetric interval,

I = \int_{- π /2}^{π /2} \frac{1 + sin ^{2} x}{1 + π ^{- s i n x}} d x = \int_{- π /2}^{π /2} \frac{( 1 + sin ^{2} x ) π ^{s i n x}}{1 + π ^{s i n x}} d x

Adding the two expressions, the denominators combine and

2 I = \int_{- π /2}^{π /2} (1 + sin^{2} x) d x = 2 \int_{0}^{π /2} (1 + sin^{2} x) d x

the last step using that

1 + sin^{2} x

is even. Therefore

I = \int_{0}^{π /2} (\frac{3}{2} - \frac{cos 2 x}{2}) d x = [\frac{3 x}{2} - \frac{sin 2 x}{4}]_{0}^{π /2} = \frac{3 π}{4}

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option