Question

Solve ${\int }_0^{\pi /4}{\tan }^{4}xdx$

Answer 1

Simplifying $\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{4}xdx$

$\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{4}xdx=\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{2}x\times {\tan }^{2}xdx$

$=\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{2}x\left({\sec }^{2}x-1\right)dx$

$=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\left({\tan }^{2}x{\sec }^{2}x-{\tan }^{2}x\right)dx$

$=\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{2}x{\sec }^{2}xdx-\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{2}xdx$

$=I_1+I_2$

$I_1=\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{2}x{\sec }^{2}xdx$

Put $\tan x=t$, then, ${\sec }^{2}xdx=dt$

$I_1=\underset{0}{\overset{\frac{\pi }{4}}{\int }}{t}^{2}dt$

$={\left[\frac{t^3}{3}\right]}_0^{\frac{\pi }{4}}$

$={\left[\frac{{\tan }^{3}x}{3}\right]}_0^{\frac{\pi }{4}}$

$=\frac{1}{3}\left({\tan }^3\frac{\pi }{4}-{\tan }^3\left(0\right)\right)$

$=\frac{1}{3}\left(1-0\right)$

$=\frac{1}{3}$

$I_2=\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{2}xdx$

$=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\left({\sec }^{2}x-1\right)dx$

$={\left[\tan x-x\right]}_0^{\frac{\pi }{4}}$

$=\tan \frac{\pi }{4}-\frac{\pi }{4}-\tan \left(0\right)+0$

$=1-\frac{\pi }{4}$

Now,

$\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{4}xdx=\frac{1}{3}+1-\frac{\pi }{4}$

$=\frac{4+12-3\pi }{12}$

$=\frac{16-3\pi }{12}$