Question

Evaluate:

${\int }_0^{\pi /4}\frac{{\tan }^{3}x}{1+\cos 2x}dx$

• A. $\frac{1}{8}$
• B. $\frac{3}{8}$
• C. $\frac{5}{8}$
• D. $\frac{7}{8}$

Answer 1

Answer: (A)

$\frac{1}{8}$

$I={\int }_0^{\pi /4}\frac{{\tan }^{3}x}{2{\cos }^{2}x}dx$

$=\frac{1}{2}{\int }_0^{\pi /4}{\tan }^{3}x{\sec }^{2}xdx$

$t=\tan x$

differentiating with respect to $x$, we get

$\frac{dt}{dx}={\sec }^{2}x\Rightarrow dt={\sec }^{2}x\cdot dx$

now, $\tan \frac{\pi }{4}=1$ and $\tan 0=0$

Replacing in the equation, we get

$\frac{1}{2}{\int }_0^1{t}^{3}dt$, where $t=\tan x$

$\Rightarrow I=\frac{1}{2}{\left[\frac{t^4}{4}\right]}_0^1=\frac{1}{2}(\frac{1}{4}-0)=\frac{1}{8}$