Question

$\int \frac{dx}{co{s}^{3}x\sqrt{sin2x}}$ $=\frac{1}{\sqrt{2}}[\frac{t^a}{a}+\frac{t^b}{b}]+C$, where $t=tanx$. Find $a+b$

Answer 1

$\int \frac{dx}{co{s}^{3}x\sqrt{sin2x}}$
$=\frac{1}{\sqrt{2}}\int \frac{dx}{co{s}^{8/2}xta{n}^{1/2}x}$
$=\frac{1}{\sqrt{2}}\int \frac{dx}{co{s}^{4}xta{n}^{1/2}x}$
$=\frac{1}{\sqrt{2}}\int \frac{(1+ta{n}^{2}x)se{c}^{2}xdx}{ta{n}^{1/2}x}$
$=\frac{1}{\sqrt{2}}\int \frac{(1+t^2)dt}{t^{1/2}}$
$=\frac{1}{\sqrt{2}}\int (t^{-1/2}+t^{3/2})dt$
$=\frac{1}{\sqrt{2}}[\frac{t^{1/2}}{1/2}+\frac{t^{5/2}}{5/2}]+C$, where $t=tanx$