Question

If $\int \frac{1}{{\cos }^{3}x\sqrt{2\sin 2x}}dx=(\tan x{)}^A+C(\tan x{)}^B+k$, where $k$ is a constant of integration, then $A+B+C$ is equal to

• A. $\frac{7}{10}$
• B. $\frac{16}{5}$
• C. $\frac{27}{10}$
• D. $\frac{21}{5}$

Answer 1

Answer: (B)

$I=\int \frac{{\sec }^{4}x}{2\sqrt{\tan x}}dx$
put $\tan x=t$
${\sec }^{2}xdx=dt$
$\Rightarrow I=\frac{1}{2}\int \frac{\left(1+t^2\right)}{\sqrt{t}}dt$
$\Rightarrow I=\frac{1}{2}\left[\frac{\sqrt{t}}{\frac{1}{2}}+\frac{t^{5/2}}{\frac{5}{2}}\right]+k$
$\Rightarrow I=(\tan x{)}^{1/2}+\frac{1}{5}(\tan x{)}^{5/2}+k$
so $A+B+C=\frac{1}{2}+\frac{1}{5}+\frac{5}{2}=\frac{5+2+25}{10}=\frac{16}{5}$