JEE PYQ: Indefinite Integration Question 27
Question 27 - 2019 (09 Jan Shift 1)
For $x^2 \neq n\pi + 1$, $n \in \mathbb{N}$ (the set of natural numbers), the integral $\int x\sqrt{\frac{2\sin(x^2-1) - \sin 2(x^2-1)}{2\sin(x^2-1) + \sin 2(x^2-1)}},dx$ is equal to:
(1) $\log_e\left|\frac{1}{2}\sec^2(x^2-1)\right| + c$
(2) $\frac{1}{2}\log_e|\sec(x^2-1)| + c$
(3) $\frac{1}{2}\log_e\left|\sec^2\left(\frac{x^2-1}{2}\right)\right| + c$
(4) $\log_e\left|\sec\left(\frac{x^2-1}{3}\right)\right| + c$
Show Answer
Answer: (3)
Solution
Inside the square root: $\frac{2\sin\theta - \sin 2\theta}{2\sin\theta + \sin 2\theta} = \frac{1-\cos\theta}{1+\cos\theta} = \tan^2\frac{\theta}{2}$ where $\theta = x^2-1$. So $I = \int x\tan\frac{x^2-1}{2},dx$. Let $\frac{x^2-1}{2} = t$, $x,dx = dt$. $I = \int \tan t,dt = \ln|\sec t| + c = \frac{1}{2}\log_e|\sec^2\frac{x^2-1}{2}| + c$.
BETA
