Question

If $I=\int \frac{\sqrt{{\sin }^{3}2x}}{{\sin }^{5}x}dx$, and $f\left(x\right)=(\cot x{)}^{3/2},g(x)=(\cot x{)}^{5/2}$, then I equals

• A. $\frac{2\sqrt{3}}{3}f\left(x\right)-\frac{g\left(x\right)}{5}+C$
• B. $-\frac{4\sqrt{2}}{5}g\left(x\right)+C$
• C. $\frac{1}{2\sqrt{3}}f\left(x\right)+C$
• D. $\frac{2\sqrt{2}}{3}f\left(x\right)+\frac{g\left(x\right)}{5}+C$

Answer 1

The correct option is B $-\frac{4\sqrt{2}}{5}g\left(x\right)+C$

$I=2\sqrt{2}\int \frac{{\left(\cos x\right)}^{3/2}}{{\left(\sin x\right)}^{7/2}}dx$
$=2\sqrt{2}\int {\left(\cot x\right)}^{3/2}cose{c}^{2}xdx$

Substitute $\cot x=t$, $-cose{c}^{2}x=dt$

we get

$I=-2\sqrt{2}\int t^{3/2}dt=-\frac{4\sqrt{2}}{5}{t}^{5/2}+c=-\frac{4\sqrt{2}}{5}{\left(\cot x\right)}^{5/2}+c$

$=-\frac{4\sqrt{2}}{5}g\left(x\right)+C$