Indefinite Integration Question 3
Question 3 - 2024 (29 Jan Shift 2)
If
$\int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^{3} x \cos ^{3} x \sin (x-\theta)}} d x=A \sqrt{\cos \theta \tan x-\sin \theta}+B \sqrt{\cos \theta-\sin \theta \cot x}+C$
where $C$ is the integration constant, then $AB$ is equal to
(1) $4 \operatorname{cosec}(2 \theta)$
(2) $4 \sec \theta$
(3) $2 \sec \theta$
(4) $8 \operatorname{cosec}(2 \theta)$
Show Answer
Answer (4)
Solution
$$ \begin{aligned} & \int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^{3} x \cos ^{3} x \sin (x-\theta)}} d x \\ & I=\int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^{3} x \cos ^{3} x(\sin x \cos \theta-\cos x \sin \theta)}} d x \\ & =\int \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x \cos ^{2} x \sqrt{\tan x \cos \theta-\sin \theta}} d x+\int \frac{\cos ^{\frac{3}{2}} x}{\sin ^{2} x \cos ^{\frac{3}{2}} x \sqrt{\cos \theta-\cot x \sin \theta}} d x= \\ & \int \frac{\sec ^{2} x}{\sqrt{\tan x \cos \theta-\sin \theta}} d x+\int \frac{\operatorname{cosec}^{2} x}{\sqrt{\cos \theta-\cot x \sin \theta}} d x \\ & I=I _1+I _2 \quad \ldots . \text { Let } \end{aligned} $$
For $I _1$, let $\tan x \cos \theta-\sin \theta=t^{2}$
$\sec ^{2} x d x=\frac{2 t d t}{\cos \theta}$
For $I _2$, let $\cos \theta-\cot x \sin \theta=z^{2}$
$\operatorname{cosec}^{2} x d x=\frac{2 z d z}{\sin \theta}$
$I=I _1+I _2$
$=\int \frac{2 t d t}{\cos \theta t}+\int \frac{2 z d z}{\sin \theta z}$
$=\frac{2 t}{\cos \theta}+\frac{2 z}{\sin \theta}$
$=2 \sec \theta \sqrt{\tan x \cos \theta-\sin \theta}+2 \operatorname{cosec} \theta \sqrt{\cos \theta-\cot x \sin \theta}$
Comparing
$A B=8 \operatorname{cosec} 2 \theta$
