JEE PYQ: Indefinite Integration Question 9
Question 9 - 2020 (04 Sep Shift 1)
The integral $\int\left(\frac{x}{x\sin x + \cos x}\right)^2,dx$ is equal to (where $C$ is a constant of integration):
(1) $\tan x - \frac{x\sec x}{x\sin x + \cos x} + C$
(2) $\sec x + \frac{x\tan x}{x\sin x + \cos x} + C$
(3) $\sec x - \frac{x\tan x}{x\sin x + \cos x} + C$
(4) $\tan x + \frac{x\sec x}{x\sin x + \cos x} + C$
Show Answer
Answer: (1)
Solution
Note $\frac{d}{dx}(x\sin x + \cos x) = x\cos x$. Write as $\int \frac{x\cos x}{(x\sin x + \cos x)^2}\cdot\frac{x}{\cos x},dx$. By parts: $= \frac{-x}{\cos x(x\sin x + \cos x)} + \int \sec^2 x,dx = -\frac{x\sec x}{x\sin x + \cos x} + \tan x + C$.
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