JEE PYQ: Indefinite Integration Question 4
Question 4 - 2021 (24 Feb Shift 1)
If $\int \frac{\cos x - \sin x}{\sqrt{8 - \sin 2x}},dx = a\sin^{-1}\left(\frac{\sin x + \cos x}{b}\right) + c$, where $c$ is a constant of integration, then the ordered pair $(a, b)$ is equal to:
(1) $(1, -3)$
(2) $(1, 3)$
(3) $(-1, 3)$
(4) $(3, 1)$
Show Answer
Answer: (2)
Solution
Put $\sin x + \cos x = t$, so $(\cos x - \sin x),dx = dt$ and $\sin 2x = t^2 - 1$. Then $I = \int \frac{dt}{\sqrt{8-(t^2-1)}} = \int \frac{dt}{\sqrt{9-t^2}} = \sin^{-1}\left(\frac{t}{3}\right) + c$. So $a = 1$, $b = 3$.
BETA
