Question

If $\int \frac{1}{\cos ^3 x \sqrt{2 \sin 2 x}} d x=(\tan x)^A+C(\tan x)^B+k$, 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)

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