JEE Main 2019 (Online) 9th January Morning ShiftQuestion 1
For $x^2\ne n\pi+1$, $n\in N$ (the set of natural numbers), the integral $\int x\sqrt{\frac{2\sin\left(x^2-1\right)-\sin2\left(x^2-1\right)}{2\sin\left(x^2-1\right)+\sin2\left(x^2-1\right)}}\,dx$ is equal to (where $c$ is a constant of integration).
Indefinite Integration
Solution
Concept. With $u=x^2-1$, the fraction under the root is $\tan^2\frac u2$. Formulas used. (1) $\dfrac{2\sin u-\sin2u}{2\sin u+\sin2u}=\dfrac{1-\cos u}{1+\cos u}=\tan^2\dfrac u2$(2) $\int\tan v\,dv=\ln|\sec v|+C$Step 1 - Simplify. By (1), the integrand is $x\tan\dfrac{x^2-1}{2}$. Step 2 - Substitute. $v=\dfrac{x^2-1}{2}$, $dv=x\,dx$. By (2), $$\int\tan v\,dv=\ln\sec\frac{x^2-1}{2}+c.$$ Answer: (D)