Q110 / 160 · AP EAMCET 2024 (20 May Shift 2) · Mathematics
$\int \frac{d x}{\sqrt{\sin ^3 x \cos (x-\alpha)}}=$
- A$\frac{1}{\sqrt{\cos \alpha}} \sqrt{\cot ^4 x+\tan \alpha}+c$
- B$\frac{1}{\sqrt{\cos \alpha}} \sqrt{\cot x-\tan \alpha}+c$
- C$\frac{-1}{\sqrt{\sin \alpha}} \sqrt{\cot x+\tan \alpha}+c$
- Correct answerD$\frac{-2}{\sqrt{\cos \alpha}} \sqrt{\cot x+\tan \alpha}+c$
Solution
$I=\int \frac{d x}{\sqrt{\sin ^3 x \cos (x-\alpha)}}$
$\begin{aligned}
& =\int \frac{d x}{\sqrt{\sin ^3 x[\cos x \cos \alpha+\sin x \sin \alpha]}} \\
& =\int \frac{d x}{\sqrt{\sin ^4 x[\cot x \cos \alpha+\sin \alpha]}} \\
& =\int \frac{d x}{\sin ^2 x \sqrt{\cos x \cos \alpha+\sin \alpha}}
\end{aligned}$
Let $\cot x \cos \alpha+\sin \alpha=t$ Differentiate w.r.t. $x,-\cos \alpha \operatorname{cosec}^2 x d x=d t$ $\begin{aligned} & \Rightarrow \mathrm{I}=\frac{1}{-\cos \alpha} \int \frac{1}{\sqrt{t}} d t \Rightarrow \mathrm{I}=\frac{-1}{\cos \alpha} 2 \sqrt{t}+\mathrm{C} \\ & \Rightarrow \mathrm{I}=\frac{-2}{\cos \alpha} \sqrt{\cot x \cos \alpha+\sin \alpha}+\mathrm{C} \\ & \mathrm{I}=\frac{-2}{\sqrt{\cos \alpha}} \sqrt{\cot x+\tan \alpha}+\mathrm{C} \end{aligned}$
Let $\cot x \cos \alpha+\sin \alpha=t$ Differentiate w.r.t. $x,-\cos \alpha \operatorname{cosec}^2 x d x=d t$ $\begin{aligned} & \Rightarrow \mathrm{I}=\frac{1}{-\cos \alpha} \int \frac{1}{\sqrt{t}} d t \Rightarrow \mathrm{I}=\frac{-1}{\cos \alpha} 2 \sqrt{t}+\mathrm{C} \\ & \Rightarrow \mathrm{I}=\frac{-2}{\cos \alpha} \sqrt{\cot x \cos \alpha+\sin \alpha}+\mathrm{C} \\ & \mathrm{I}=\frac{-2}{\sqrt{\cos \alpha}} \sqrt{\cot x+\tan \alpha}+\mathrm{C} \end{aligned}$
Hence, the correct answer is option: 'D'.