Question

If $\int \frac{dx}{\sqrt{{\sin }^{3}x{\cos }^{5}x}}=a\sqrt{\cot x}+b\sqrt{{\tan }^{3}x}+C$, then

• A. $a=-1,b=\frac{1}{3}$
• B. $a=-3,b=\frac{2}{3}$
• C. $a=-2,b=\frac{4}{3}$
• D. $a=-2,b=\frac{2}{3}$

Answer 1

The correct option is D $a=-2,b=\frac{2}{3}$

Let $I=\frac{dx}{\sqrt{\frac{{\sin }^{3}x}{{\cos }^{3}x}{\cos }^{8}x}}$
$=\int \frac{{\sec }^{4}x}{\sqrt{{\tan }^{3}x}}dx$
$=\int \frac{(1+{\tan }^{2}x){\sec }^{2}xdx}{\sqrt{{\tan }^{3}x}}$
Put $\tan x=t$
$\Rightarrow {\sec }^{2}xdx=dt$
So, $I=\int \frac{1+t^2}{t^{3/2}}dt$
$=\int t^{-3/2}dt+\int t^{1/2}dt$
$=-2t^{-1/2}+\frac{2}{3}{t}^{3/2}$
$=-2\sqrt{\cot x}+\frac{2}{3}\sqrt{{\tan }^{3}x}+C$
Hence, $-2\sqrt{\cot x}+\frac{2}{3}\sqrt{{\tan }^{3}x}+C=a\sqrt{\cot x}+b\sqrt{{\tan }^{3}x}+C$
$\Rightarrow a=-2,b=\frac{2}{3}$