Question

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

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

Answer 1

The correct option is C $\frac{16}{5}$

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

$\Rightarrow I=\int \frac{{\sec }^{3}x}{2\sqrt{\sin x\cos x}}dx$

Multiplying numerator and and denominator of $I$ by $\sec x$, we get

$I=\int \frac{{\sec }^{4}x}{2\sqrt{\tan x}}dx$

Now assuming, $\tan x=t^2\Rightarrow {\sec }^{2}xdx=2tdt$, we get

$I=\int \frac{1+t^4}{2t}\left(2t\right)dt$

$\Rightarrow I=\int (1+t^4)dt$

$\Rightarrow I=t+\frac{t^5}{5}+K$

Re-Substituting $t=\sqrt{\tan x}$, we get

$I=\sqrt{\tan x}+\frac{1}{5}\cdot {\sqrt{\tan x}}^5+K$

Comparing it with the given expression

$I=\sqrt{\tan x}+\frac{1}{5}\cdot {\sqrt{\tan x}}^5+K$ $=(\tan x{)}^A+C(\tan x{)}^B+K$

$\Rightarrow A=\frac{1}{2},B=\frac{5}{2},C=\frac{1}{5}$

$A+B+C=\frac{1}{2}+\frac{5}{2}+\frac{1}{5}=\frac{16}{5}$