Question

If $\int \frac{{\sin }^{3}x}{{\cos }^{5}x}dx=\frac{\left(f\right(x){)}^4}{4}+C$ and $\int \left(f\right(x){)}^{5}dx=\frac{\left(f\right(x){)}^4}{4}-\frac{\left(f\right(x){)}^2}{2}+\ln |g(x)|,$ then $g\left(\frac{\pi }{3}\right)=$
(where $C$ is integration constant)

• A. $2$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{1}{2}$
• D. $\frac{2}{\sqrt{3}}$

Answer 1

The correct option is A $2$

$\int \frac{{\sin }^{3}x}{{\cos }^{5}x}dx=\int {\tan }^{3}x\cdot {\sec }^{2}xdx=\frac{{\tan }^{4}x}{4}+C$
$\therefore f\left(x\right)=\tan x$
$\int {\tan }^{5}xdx=\frac{{\tan }^{4}x}{4}-\int {\tan }^{3}xdx=\frac{{\tan }^{4}x}{4}-\frac{{\tan }^{2}x}{2}+\int \tan xdx=\frac{{\tan }^{4}x}{4}-\frac{{\tan }^{2}x}{2}+\ln |\sec x|+C^{\prime }=\frac{{\tan }^{4}x}{4}-\frac{{\tan }^{2}x}{2}+\ln |\sec x|(\because C^{\prime }=0)$ (given)
So, $g\left(x\right)=\sec x\Rightarrow g\left(\frac{\pi }{3}\right)=2$