For $x^{2} \neq n \pi+1, n \in N$ (the set of natural numbers), the integral $\int x \cdot \sqrt{\frac{2 \sin \left(x^{2}-1\right)-\sin 2\left(x^{2}-1\right)}{2 \sin \left(x^{2}-1\right)+\sin 2\left(x^{2}-1\right)}} d x$ is

(D) $\log _{e}\left|\sec \left(\frac{x^{2}-1}{2}\right)\right|+c$

(A) $\log _{e}\left|\frac{1}{2} \sec ^{2}\left(x^{2}-1\right)\right|+c$

(B) $\frac{1}{2} \log _{e}\left|\sec \left(x^{2}-1\right)\right|+c$

(C) $\frac{1}{2} \log _{e}\left|\sec ^{2}\left(\frac{x^{2}-1}{2}\right)\right|+c$

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Indefinite Integrals

For

x^{2} \neq = nπ + 1, n \in N

(the set of natural numbers), the integral

\int x \cdot \frac{2 s i n ( x ^{2} - 1 ) - s i n 2 ( x ^{2} - 1 )}{2 s i n ( x ^{2} - 1 ) + s i n 2 ( x ^{2} - 1 )} d x

Answer: (D)

Rewrite the ratio under the root with

sin 2 α = 2 sin α cos α

, where

α = x^{2} - 1

: it becomes

\frac{2 s i n α ( 1 - c o s α )}{2 s i n α ( 1 + c o s α )} = \frac{1 - c o s α}{1 + c o s α} = tan \frac{α}{2}

. The integral is then

\int x tan \frac{x ^{2} - 1}{2} d x

. Substituting

t = \frac{x ^{2} - 1}{2}

so that

d t = x d x

turns it into

\int tan t d t = ln ∣ sec t ∣ + c = ln sec \frac{x ^{2} - 1}{2} + c

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option