The value of the integral $\int_{0}^{\frac{\pi}{4}} \frac{x dx}{\sin ^{4}(2 x)+\cos ^{4}(2 x)}$ equals :

(C) $\frac{\sqrt{2} \pi^{2}}{32}$

(A) $\frac{\sqrt{2} \pi^{2}}{8}$

(B) $\frac{\sqrt{2} \pi^{2}}{16}$

(D) $\frac{\sqrt{2} \pi^{2}}{64}$

Definite Integrals — JEE Main · 2024 · 01 Feb 2024 · Shift 1 Question with Solution

The value of the integral

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

equals :

Answer: (C)

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

Using

\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x

with

a = π /4

: substituting

x \to π /4 - x

sends

2 x \to π /2 - 2 x

, which swaps

sin 2 x

and

cos 2 x

and leaves the denominator unchanged, while the numerator becomes

π /4 - x

I = \int_{0}^{π /4} \frac{( π /4 - x ) d x}{sin ^{4} 2 x + cos ^{4} 2 x} = \frac{π}{4} K - I, K = \int_{0}^{π /4} \frac{d x}{sin ^{4} 2 x + cos ^{4} 2 x}

⟹ 2 I = \frac{π}{4} K

Now

sin^{4} θ + cos^{4} θ = 1 - \frac{1}{2} sin^{2} 2 θ

; with

θ = 2 x

, the denominator is

1 - \frac{1}{2} sin^{2} 4 x

. Substituting

t = 4 x

K

K = \frac{1}{4} \int_{0}^{π} \frac{d t}{1 - \frac{1}{2} sin ^{2} t}

Using the standard result

\int_{0}^{π} \frac{d t}{1 - k sin ^{2} t} = \frac{π}{1 - k}

with

k = \frac{1}{2}

K = \frac{1}{4} \cdot \frac{π}{1/2} = \frac{1}{4} \cdot π 2 = \frac{π 2}{4}

2 I = \frac{π}{4} \cdot \frac{π 2}{4} = \frac{2 π ^{2}}{16} ⟹ I = \frac{2 π ^{2}}{32}

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Single correct option