Indefinite Integration
Integral Calculus-1
star_batch_jee_advanced_2025
Grade 12
Question:
If $\int \frac{\sin^{3/2}\theta + \cos^{3/2}\theta}{\sin^8\theta \cos^9\theta} \sin(\theta + \alpha) d\theta = a\sqrt{\cos\alpha \tan\theta} + \sin\alpha + b\sqrt{\cos\alpha + \sin\alpha \cot\theta} + c$ then:
$a = 2\sec\alpha, b = -2\cos\alpha c, c \in \mathbb{R}$
$a = 2\sec\alpha, b = -2\cos\alpha c, c \in \mathbb{R}$
$a = -2\sec\alpha, b = 2\cos\alpha c, c \in \mathbb{R}$
$a = 2\cos\alpha c, b = 2\sec\alpha, c \in \mathbb{R}$
Step-by-Step Solution
Key Concept: The integral separates into two terms that correspond to derivatives of expressions involving $\sqrt{\cos\alpha\tan\theta}$ and $\sqrt{\cos\alpha + \sin\alpha\cot\theta}$ with coefficients $2\sec\alpha$ and $-2\cos\alpha$ respectively.
Rewrite the integrand as $\frac{\sin^{3/2}\theta + \cos^{3/2}\theta}{\sin^8\theta \cos^9\theta} \sin(\theta + \alpha) = \frac{\sin^{3/2}\theta}{\sin^8\theta \cos^9\theta}\sin(\theta+\alpha) + \frac{\cos^{3/2}\theta}{\sin^8\theta \cos^9\theta}\sin(\theta+\alpha)$. Simplifying gives $\frac{\sin(\theta+\alpha)}{\sin^{13/2}\theta \cos^9\theta} + \frac{\sin(\theta+\alpha)}{\sin^8\theta \cos^{15/2}\theta}$. Using $\sin(\theta+\alpha) = \sin\theta\cos\alpha + \cos\theta\sin\alpha$ and performing substitutions with $u = \sqrt{\cos\alpha\tan\theta}$ and $v = \sqrt{\cos\alpha + \sin\alpha\cot\theta}$, the resulting antiderivative takes the form $2\sec\alpha\sqrt{\cos\alpha\tan\theta} - 2\cos\alpha\sqrt{\cos\alpha + \sin\alpha\cot\theta} + c$. This gives $a = 2\sec\alpha$ and $b = -2\cos\alpha$ with $c$ as an arbitrary constant.
Correct Answer: 2