Question

The value of $\int \frac{\sqrt{{\sin }^{3}2x}}{{\sin }^{5}x}dx$ is

• A. $2^{\frac{5}{2}}\frac{{\tan }^{-\frac{5}{2}}x}{5}+C$
• B. $2^{\frac{5}{2}}\frac{{\tan }^{\frac{5}{2}}x}{5}+C$
• C. $-2^{\frac{5}{2}}\frac{{\tan }^{-\frac{5}{2}}x}{5}+C$
• D. $-2^{\frac{5}{2}}\frac{{\tan }^{\frac{5}{2}}x}{5}+C$

Answer 1

Answer: (C)

$-2^{\frac{5}{2}}\frac{{\tan }^{-\frac{5}{2}}x}{5}+C$

Consider the given integral.

$I=\int \frac{\sqrt{{\sin }^{3}2x}}{{\sin }^{5}x}dx$

$I=\int \frac{{\sin }^{\frac{3}{2}}2x}{{\sin }^{5}x}dx$

$I=\int \frac{{\left(2\sin x\cos x\right)}^{\frac{3}{2}}}{{\sin }^{5}x}dx$

$I=2^{\frac{3}{2}}\int \frac{{\sec }^{2}x}{{\tan }^{\frac{7}{2}}x}dx$

Put $u=\tan x$ $\Rightarrow$ $du={\sec }^{2}xdx$

Therefore,

$I=2^{\frac{3}{2}}\int \frac{1}{u^{\frac{7}{2}}x}du$

$I=2^{\frac{3}{2}}\int u^{-\frac{7}{2}}du$

$I=2^{\frac{3}{2}}\frac{u^{-\frac{5}{2}}}{-\frac{5}{2}}+C$

Put the value of $u$ in the above expression, we get

$I=-2^{\frac{5}{2}}\frac{{\tan }^{-\frac{5}{2}}x}{5}+C$

Hence, this is the correct answer.