Reduction Formula If $\displaystyle I_{n}\equiv
\int^{\pi/4}_{0} \tan^{n}\left(\,\theta\,\right)\,{\rm d}\theta$ prove that $\displaystyle I_{n} = {1 \over n - 1} - I_{n - 2}$
I tried using integration by parts by first writing $tan\theta$ as $tan\theta^{n-1}  tan\theta$ and then used the by parts formula but realized that it wouldn't work as $ln$ was coming into play.
I then tried it by writing $tan\theta$ as $tan\theta^{n-2}tan\theta^{2}$ and then putting $sec\theta^2 -1$ in place of $tan\theta^2$. I tried to simplify the result, getting $I_{n-2}$ in the process but failed to arrive at the given result. 
 A: By using the substitution $\theta=\arctan u$, we have:
$$ I_n = \int_{0}^{1}\frac{u^n}{u^2+1}\,du = \int_{0}^{1}u^{n-2}\,du-\int_{0}^{1}\frac{u^{n-2}}{u^2+1}\,du = \frac{1}{n-1}-I_{n-2}.$$
A: Not a complete solution, but the beginning of one.
Since the thing you need to prove is $I_n + I_{n-2} = 1/(n-1)$, it makes sense to try to calculate $I_n + I_{n-2}$. We have
$$ I_n + I_{n-2} = \int_0^{\pi/4} \tan^{n-2}\theta(\tan^2 \theta + 1) \, d\theta
= \int_0^{\pi/4} \tan^{n-2}\theta \sec^2 \theta \, d\theta. $$
Since you have $\sec^2 \theta \, d\theta$ as a factor, you can evaluate the last integral by the substitution $u = \tan \theta$.
A: It sounds like your approach was correct
$$\int_0^\frac\pi4\tan^n\theta d\theta=\int_0^\frac\pi4\tan^{n-2}\theta\sec^2\theta d\theta-\int_0^\frac\pi4\tan^{n-2}\theta d\theta=$$
$$\frac{\tan^{n-1}\theta}{n-1}|_0^\frac\pi4-I_{n-2}=\frac{1-0}{n-1}-I_{n-2}=\frac1{n-1}-I_{n-2}$$
A: $$ I_n =\int_0^{\frac{\pi}{4}} \tan^n x dx  =\int_0^{\frac{\pi}{4}}\tan^{n-2}  x (1+\tan^2 x) dx -I_{n-2} =\int_0^{1} u^{n-2} du -I_{n-2} =\frac{1}{n-1} -I_{n-2} $$ where $u=\tan x.$
