Question

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

• A. $a=-2,b=\frac{2}{3}$
• B. $a=-3,b=\frac{2}{3}$
• C. $a=-2,b=\frac{4}{3}$
• D. None of these

Answer 1

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

$I=\int \frac{dx}{\sqrt{{\sin }^{3}x{\cos }^5}\times \frac{{\cos }^{3}x}{{\cos }^{3}x}}$

$=\int \frac{dx}{\sqrt{\frac{{\sin }^{3}x}{{\cos }^{3}x}}{\cos }^{8}x}$

$=\int \frac{{\sec }^{4}x}{\sqrt{{\tan }^{3}x}}dx$

$=\int \frac{(1+{\tan }^{2}x){\sec }^{2}x}{\sqrt{{\tan }^{3}x}}dx$

$\tan x=t$

${\sec }^{2}x=\frac{dt}{dx}$

$=\int \frac{1+t^2}{t^{3/2}}dt$

$=\int (t^{-3/2}+t^{1/2})dt$

$=-1t^{-1/2}+2/3t^{3/2}+C$

$=-2\sqrt{\cot x}+2/3\sqrt{{\tan }^{3}x}+C$

$(a=-2,b=2/3)$