If \(\int_{-\pi / 2}^{\pi / 2} \frac{8 \sqrt{2} \cos x d x}{\left(1+e^{\sin x}\right)\left(1+\sin ^{4} x\right)}=\alpha \pi+\beta \log _{e}(3+2\sqrt{2})\), where \(\alpha, \beta\) are integers, then \(\alpha^{2}+\beta^{2}\) equals ______.

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

\int_{- π /2}^{π /2} \frac{8 2 c o s x d x}{( 1 + e ^{s i n x} ) ( 1 + s i n ^{4} x )} = α π + β lo g_{e} (3 + 22)

, where

α, β

are integers, then

α^{2} + β^{2}

equals ______.

Answer: 8

I = \int_{- π /2}^{π /2} \frac{8 2 cos x d x}{( 1 + e ^{s i n x} ) ( 1 + sin ^{4} x )}

Write

g (x) = \frac{8 2 cos x}{1 + sin ^{4} x}

, which is even. Using the identity

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

with

u = sin x

: applying

x \to - x

I

turns

sin x \to - sin x

(and leaves

g (x)

unchanged since it's even), giving a second expression for

I

; adding the two:

2 I = \int_{- π /2}^{π /2} g (x) [\frac{1}{1 + e ^{s i n x}} + \frac{1}{1 + e ^{- s i n x}}] d x = \int_{- π /2}^{π /2} g (x) d x = 162 \int_{0}^{π /2} \frac{cos x d x}{1 + sin ^{4} x}

⟹ I = 82 \int_{0}^{π /2} \frac{cos x d x}{1 + sin ^{4} x}

Substitute

t = sin x

I = 82 \int_{0}^{1} \frac{d t}{1 + t ^{4}}

Using the standard antiderivative

\int \frac{d t}{1 + t ^{4}} = \frac{1}{4 2} ln \frac{t ^{2} - 2 t + 1}{t ^{2} + 2 t + 1} + \frac{1}{2 2} [tan^{- 1} (2 t + 1) + tan^{- 1} (2 t - 1)] + C

:

At

t = 1

: the log ratio is

\frac{2 - 2}{2 + 2} = 3 - 22 = \frac{1}{3 + 2 2}

, so

ln (3 - 22) = - ln (3 + 22)

; the arctan sum is

tan^{- 1} (2 + 1) + tan^{- 1} (2 - 1) = \frac{3 π}{8} + \frac{π}{8} = \frac{π}{2}

.
At

t = 0

: log ratio

= 1

(

ln 1 = 0

); arctan sum

= tan^{- 1} 1 + tan^{- 1} (- 1) = \frac{π}{4} - \frac{π}{4} = 0

\int_{0}^{1} \frac{d t}{1 + t ^{4}} = \frac{1}{4 2} [- ln (3 + 22)] + \frac{1}{2 2} \cdot \frac{π}{2} = \frac{1}{4 2} [π - ln (3 + 22)]

I = 82 \cdot \frac{1}{4 2} [π - ln (3 + 22)] = 2 π - 2 ln_{e} (3 + 22)

Matching

I = α π + β ln_{e} (3 + 22)

α = 2, β = - 2

, so

α^{2} + β^{2} = 4 + 4 = 8

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Integer answer