# Thread: How do I finish this evaluation of reduction formula?

1. ## How do I finish this evaluation of reduction formula?

I wanted to find a reduction formula for $\displaystyle \int_0^{\pi/4} tan^n x dx$

And I found this as $\displaystyle I_{n} = \frac{1}{n-1} - I_{n-2}$

I found $\displaystyle I_1$ as $\displaystyle \frac{ln2}{2}$ and the last thing I want to do is find, for example, $\displaystyle I_5$.

My working here is as follows:

$\displaystyle I_5 = \frac{1}{4} \int_0^{\pi/4} tan^3 x dx$
$\displaystyle I_3 = \frac{1}{2} \int_0^{\pi/4} tan x dx$

Which gave me

$\displaystyle I_5 = \frac{ln2}{8} - \frac{1}{8}$

Now I think that's wrong. I know I'm looking for an answer in the form

$\displaystyle a\ln2 - b$ where a and b are fractions, but I know my answer's wrong because it is negative - and I checked on wolfram alpha which gives a different integral. Could you please tell me what I have done wrong? Thanks if you can help me

2. it's actually $\displaystyle I_{n}=\frac{1}{n-1}-I_{n-2}.$

3. Originally Posted by Krizalid
it's actually $\displaystyle I_{n}=\frac{1}{n-1}-I_{n-2}.$
That was a typo there sorry! I've edited it now. By the way, everything I have done in the rest of the question was based on it being 2 rather than 1 - I know that the ln2/2 part is correct, I know it's after that where my answer falls apart.

4. $\displaystyle I_5 = \frac{1} {4} - I_3 = \frac{1} {4} - \left[ {\frac{1} {2} - I_1 } \right] = - \frac{1} {2} + I_1$

5. Originally Posted by Plato
$\displaystyle I_5 = \frac{1} {4} - I_3 = \frac{1} {4} - \left[ {\frac{1} {2} - I_1 } \right] = - \frac{1} {2} + I_1$
I am an idiot. I've just figured out what I've done. I had what you gave as the last stage, but instead of adding and subtracting the terms, I've inexplicably multiplied it all out.

Everyone thanks very much for your time