The value of the integral $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} dx$ is :

(D) $\frac{\pi^{2}}{6 \sqrt{3}}$

(B) $\frac{\pi^{2}}{12 \sqrt{3}}$

(C) $\frac{\pi^{2}}{3 \sqrt{3}}$

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Definite Integrals

The value of the integral

\int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{x + \frac{π}{4}}{2 - c o s 2 x} d x

is :

Answer: (D)

Let

I = \int_{- π /4}^{π /4} \frac{x + π /4}{2 - cos 2 x} d x

. Replacing

x

- x

keeps the denominator fixed and turns the numerator into

- x + π /4

; adding the two forms cancels the odd part, leaving

2 I = \frac{π}{2} \int_{- π /4}^{π /4} \frac{d x}{2 - cos 2 x}

, so

I = \frac{π}{2} \int_{0}^{π /4} \frac{d x}{2 - cos 2 x}

by evenness. With

t = tan x

2 - cos 2 x = \frac{1 + 3 t ^{2}}{1 + t ^{2}}

and

d x = \frac{d t}{1 + t ^{2}}

, so the integral becomes

\int_{0}^{1} \frac{d t}{1 + 3 t ^{2}} = \frac{1}{3} tan^{- 1} 3 = \frac{π}{3 3}

. Hence

I = \frac{π}{2} \cdot \frac{π}{3 3} = \frac{π ^{2}}{6 3}

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Single correct option