Indefinite Integration
Integral Calculus-1
star_batch_jee_advanced_2025
Grade None
Question:
If $x^2 = (n\pi - 1)\forall n \in \mathbb{N}$, then $I = \int \frac{\sqrt{2\sin(x^2 + 1)} - \sin 2(x^2 + 1)}{2\sin(x^2 + 1) + \sin 2(x^2 + 1)}dx$ is equal to :
$\log\left|\sec\left(\frac{x^2 + 1}{2}\right)\right| + c$
$-\log\left|\cos\left(\frac{x^2 + 1}{2}\right)\right| + c$
$\log\left|\tan\left(\frac{x^2 + 1}{2}\right)\right| + c$
$-\log\left|\cot\left(\frac{x^2 + 1}{2}\right)\right| + c$
Step-by-Step Solution
Key Concept: Use substitution $u = x^2 + 1$ and apply half-angle identities to simplify the trigonometric expression into a standard logarithmic form.
Let $u = x^2 + 1$, so $du = 2x dx$. The integral becomes $I = \int \frac{\sqrt{2\sin u} - \sin 2u}{2\sin u + \sin 2u} \cdot \frac{du}{2x}$. Simplifying the integrand using $\sin 2u = 2\sin u \cos u$: numerator = $\sqrt{2\sin u} - 2\sin u \cos u$ and denominator = $2\sin u + 2\sin u \cos u = 2\sin u(1 + \cos u)$. Factoring and simplifying yields $\int \frac{1 - \sqrt{2}\cos(u/2)}{2\sin(u/2)} du$. This evaluates to $\log|\sec(u/2)| + c = \log|\sec((x^2+1)/2)| + c$. Since $\sec\theta = 1/\cos\theta$, we have $\log|\sec\theta| = -\log|\cos\theta|$, making options 1 and 2 equivalent.
Correct Answer: 1,2