Integration problem

[Yankel]

Hello all,

I am trying to integrate the function

f(x)=tan^5(x)/cos^2(x)

I can't figure out how to do it

the answer should be tan^6(x)/6 + C

thanks...

 

[Random Variable]

If you make the substitution $u = \tan x$, then $ \displaystyle du = \sec^{2} x \ dx = \frac{1}{\cos^{2} x} \ dx $.

 

[Yankel]

Thank you, your tip was great help.

I want to ask for help with one more integral, seems a waste to start a new thread, I need this:

\[ln(x^{2}-1)\]

I tried using integration in parts and got stuck with the integral of:

\[\frac{2x^{2}}{x^{2}-1}\]

Thanks !

 

[Random Variable]

Using polynomial long division and then the method of partial fractions,

$$\frac{x^{2}}{x^{2}-1} = 1 + \frac{1}{x^{2}-1} = 1 + \frac{1}{2} \Big( \frac{1}{x-1} - \frac{1}{x+1} \Big)$$

 

[Ackbach]

I want to ask for help with one more integral, seems a waste to start a new thread,

Actually, that's exactly what we do prefer here at MHB. You can ask two questions per thread, but please do not start a new question in the middle of an existing thread.

Thanks!

 

[Krizalid]

I want to ask for help with one more integral, seems a waste to start a new thread, I need this:

\[ln(x^{2}-1)\]

I tried using integration in parts and got stuck with the integral of:

\[\frac{2x^{2}}{x^{2}-1}\]

Thanks !

\(\displaystyle \ln(x^2-1)=\ln(x+1)+\ln(x-1).\)