The integral $\int_{0}^{\pi} \frac{8 x d x}{4 \cos ^{2} x+\sin ^{2} x}$ is equal to

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Definite Integrals

The integral

\int_{0}^{π} \frac{8 x d x}{4 c o s ^{2} x + s i n ^{2} x}

is equal to

Answer: (A)

Let

I = \int_{0}^{π} \frac{8 x d x}{4 c o s ^{2} x + s i n ^{2} x}

. Replacing

x

π - x

leaves the denominator unchanged, so

I = \int_{0}^{π} \frac{8 ( π - x ) d x}{4 c o s ^{2} x + s i n ^{2} x}

; adding,

2 I = 8 π \int_{0}^{π} \frac{d x}{4 c o s ^{2} x + s i n ^{2} x}

. The denominator is symmetric about

x = \frac{π}{2}

, so this is

16 π \int_{0}^{π /2} \frac{s e c ^{2} x d x}{4 + t a n ^{2} x}

. With

t = tan x

16 π \int_{0}^{\infty} \frac{d t}{4 + t ^{2}} = 16 π \cdot \frac{1}{2} \cdot \frac{π}{2} = 4 π^{2}

, so

2 I = 4 π^{2}

and the value of the integral is

2 π^{2}

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Single correct option