Indefinite Integration
Integral Involving sin and cos Powers — Substitution
nta_pyq_2024_jan
Grade 12
Question:
If $\displaystyle\int\dfrac{\sin^{3/2}x+\cos^{3/2}x}{\sqrt{\sin^3x\cos^3x\sin(x-\theta)}}\,dx=A\sqrt{\cos\theta\tan x-\sin\theta}+B\sqrt{\cos\theta-\sin\theta\cot x}+C$, where $C$ is the integration constant, then $AB$ is equal to
$4\csc(2\theta)$
$4\sec\theta$
$2\sec\theta$
$8\csc(2\theta)$
Step-by-Step Solution
Key Concept: Split the integral as $I=I_1+I_2$. For $I_1$: numerator $\sin^{3/2}x$, let $u=\tan x\cos\theta-\sin\theta$. For $I_2$: numerator $\cos^{3/2}x$, let $v=\cos\theta-\cot x\sin\theta$. Integrate each using $\int\frac{du}{\sqrt{u}}=2\sqrt{u}$.
$I_1=\int\frac{\sec^2x}{\sqrt{\tan x\cos\theta-\sin\theta}}dx=\frac{2t}{\cos\theta}\Big|_{t=\sqrt{\tan x\cos\theta-\sin\theta}}=\frac{2}{\cos\theta}\sqrt{\tan x\cos\theta-\sin\theta}$. Similarly $I_2=\frac{2}{\sin\theta}\sqrt{\cos\theta-\cot x\sin\theta}$. $A=2\sec\theta,B=2\csc\theta$. $AB=4\sec\theta\csc\theta=8\csc(2\theta)$.
Correct Answer: 4