Practicing Success
If $I_n=\int\limits_0^{\pi / 4} \tan ^n x d x, n \in N$, then $I_{n+2}+I_n$ equals |
$\frac{1}{n}$ $\frac{1}{n-1}$ $\frac{1}{n+1}$ $\frac{1}{n+2}$ |
$\frac{1}{n+1}$ |
We have, $I_{n+2}+I_n=\int\limits_0^{\pi / 4} \tan ^{n+2} x d x+\int\limits_0^{\pi / 4} \tan ^n x d x$ $\Rightarrow I_{n+2}+I_n=\int\limits_0^{\pi / 4} \tan ^n x\left(1+\tan ^2 x\right) d x=\int\limits_0^{\pi / 4} \tan ^n x \sec ^2 x d x$ $\Rightarrow I_{n+2}+I_n=\int\limits_0^1 t^n d t$, where $t=\tan x$ $\Rightarrow I_{n+2}+I_n=\frac{1}{n+1}$ |