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

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Definite Integrals

The value of

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

is equal to:

Answer: (C)

Let

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

. Replacing

x

2 π - x

and adding the result to

I

gives

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

. Over the full period the integrand pairs with its

sin \leftrightarrow cos

complement to sum to

1

, so that integral equals

π

. Thus

2 I = 2 π \cdot π = 2 π^{2}

, giving

I = π^{2}

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Single correct option

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