If $\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x=a \sin ^{-1}\left(\frac{\sin x+\cos x}{b}\right)+c$, where $c$ is a constant of integration, then the ordered pair ( $a, b$ ) is equal to:

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Indefinite Integrals

\int \frac{c o s x - s i n x}{8 - s i n 2 x} d x = a sin^{- 1} (\frac{s i n x + c o s x}{b}) + c

, where

c

is a constant of integration, then the ordered pair (

a, b

) is equal to:

Answer: (D)

Let

t = sin x + cos x

. Then

d t = (cos x - sin x) d x

and

t^{2} = 1 + sin 2 x

, so

sin 2 x = t^{2} - 1

.

The integral becomes:

\int \frac{d t}{8 - ( t ^{2} - 1 )} = \int \frac{d t}{9 - t ^{2}} = sin^{- 1} (\frac{t}{3}) + c

Substituting back:

sin^{- 1} (\frac{s i n x + c o s x}{3}) + c

Comparing with

a sin^{- 1} (\frac{s i n x + c o s x}{b}) + c

, we get

a = 1

and

b = 3

.

Therefore,

(a, b) = (1, 3)

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Single correct option