The integral $\int_{0}^{\pi} \frac{(x+3) \sin x}{1+3 \cos ^{2} x} d x$ is equal to :

(C) $\frac{\pi}{3 \sqrt{3}}(\pi+6)$

(A) $\frac{\pi}{\sqrt{3}}(\pi+1)$

(B) $\frac{\pi}{\sqrt{3}}(\pi+2)$

(D) $\frac{\pi}{2 \sqrt{3}}(\pi+4)$

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Definite Integrals

The integral

\int_{0}^{π} \frac{( x + 3 ) s i n x}{1 + 3 c o s ^{2} x} d x

is equal to :

Answer: (C)

Replacing

x

π - x

I = \int_{0}^{π} \frac{( x + 3 ) sin x}{1 + 3 cos ^{2} x} d x

and adding the two forms removes the variable part of the numerator, giving

2 I = \int_{0}^{π} \frac{( π + 6 ) sin x}{1 + 3 cos ^{2} x} d x

. The substitution

u = cos x

turns this into

\int_{- 1}^{1} \frac{d u}{1 + 3 u ^{2}} = \frac{2}{3} tan^{- 1} 3 = \frac{2}{3} \cdot \frac{π}{3} = \frac{2 π}{3 3}

. Hence

2 I = (π + 6) \cdot \frac{2 π}{3 3}

, so

I = \frac{π}{3 3} (π + 6)

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Single correct option