# Integration Problem

1. Jan 15, 2007

### Xaif

There's a question here from a past exam paper I don't understand. I have the answer that my lecturer gave but the problem is I just don't really understand it. I'm hoping that another explanation from a different person may help. The question is:

I'd appreciate any feedback, thanks.

2. Jan 15, 2007

### benorin

Let $$I_{2n}=\int_{0}^{\frac{\pi}{4}}\tan^{2n}x\, dx$$, then

$$I_{2n}=\int_{0}^{\frac{\pi}{4}}\tan^{2n-2}x(\sec^{2}x-1)\, dx=\int_{0}^{\frac{\pi}{4}}\tan^{2n-2}x\sec^{2}x\, dx-\int_{0}^{\frac{\pi}{4}}\tan^{2n-2}x\, dx=\int_{0}^{\frac{\pi}{4}}\tan^{2n-2}x\sec^{2}x\, dx-I_{2n-2}$$​

now substitute $$u=\tan x\Rightarrow du=\sec^{2}x\,dx$$ so that $$0\leq x\leq \frac{\pi}{4}\Rightarrow 0\leq u\leq 1$$ and the integral becomes

$$I_{2n}=\int_{0}^{1}u^{2n-2}du-I_{2n-2}=\left[\frac{u^{2n-1}}{2n-1}\right]_{u=0}^{1}-I_{2n-2}=\frac{1}{2n-1}-I_{2n-2}$$​