The value of the definite integral $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\left(1+e^{x \cos x}\right)\left(\sin ^{4} x+\cos ^{4} x\right)}$ is equal to :

Definite Integrals — JEE Main · 2021 · 27 Jul 2021 · Shift 1 Question with Solution

The value of the definite integral

\int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{d x}{( 1 + e ^{x c o s x} ) ( s i n ^{4} x + c o s ^{4} x )}

is equal to :

Answer: (B)

Write

g (x) = sin^{4} x + cos^{4} x

, which is even, and let

I

be the given integral. Replacing

x

- x

and using

\frac{1}{1 + e ^{- u}} = \frac{e ^{u}}{1 + e ^{u}}

with the odd function

u = x cos x

I = \int_{- π /4}^{π /4} \frac{e ^{x c o s x}}{( 1 + e ^{x c o s x} ) g ( x )} d x

Adding this to the original form,

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

Dividing numerator and denominator by

cos^{4} x

and putting

t = tan x

\int_{0}^{π /4} \frac{sec ^{4} x d x}{tan ^{4} x + 1} = \int_{0}^{1} \frac{1 + t ^{2}}{1 + t ^{4}} d t = \int_{0}^{1} \frac{1 + \frac{1}{t ^{2}}}{t ^{2} + \frac{1}{t ^{2}}} d t

With

u = t - \frac{1}{t}

, so that

d u = (1 + \frac{1}{t ^{2}}) d t

and

t^{2} + \frac{1}{t ^{2}} = u^{2} + 2

, the limits

t = 0^{+}

and

t = 1

become

u = - \infty

and

u = 0

\int_{- \infty}^{0} \frac{d u}{u ^{2} + 2} = \frac{1}{2} [tan^{- 1} \frac{u}{2}]_{- \infty}^{0} = \frac{π}{2 2}

Hence

2 I = 2 \cdot \frac{π}{2 2}

, so the integral equals

\frac{π}{2 2}

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option