Question

If $I_n=\stackrel{\pi /4}{\underset{0}{\int }}{\tan }^{n}xdx$, then prove that $(n+2014)(I_{n+2013}+I_{n+2015})=1$.

Answer 1

$I_n=\stackrel{\pi /4}{\underset{0}{\int }}{\tan }^{n}xdx$
$\therefore I=(n+2014)[I_{(n+2013)}+I_{(n+2015)}]$
$=(n+2014)\left\{\overset{\pi .4}{\underset{0}{\displaystyle \int}}\tan^{n+2013}xdx+\overset{\pi /4}{\underset{0}{\displaystyle\int}}\tan^{n+2015}xdx\right\}$
$=(n+2014)\left\{\overset{\pi /4}{\underset{0}{\displaystyle\int}}(\tan^{n+2013}x+\tan^{n+2015}x)\right\}$
$=(n+2014)\stackrel{\pi /4}{\underset{0}{\int }}{\tan }^{n+2013}x[1+ta{n}^{2}x]dx$
$=(n+2014)\stackrel{\pi /4}{\underset{0}{\int }}{\tan }^{n+2013}x\cdot {\sec }^{2}xdx$
Now taking $\tan x=t$
$\therefore se{c}^{2}xdx=dt$
If $x=\frac{\pi }{4}$ then $t=\tan \frac{\pi }{4}$
$\therefore t=1$
and if $x=0$ then $t=\tan 0$
$\therefore t=0$
$\therefore I=(n+2014)\stackrel{1}{\underset{0}{\int }}{t}^{n+2013}\cdot dt$
$=(n+2014){\left[\frac{t^{n+2014}}{n+2014}\right]}_0^1$
$=(1{)}^{n+2014}=1$
$=1$
$\therefore I=1$.