Integrate f(x)=tan^5(x)/cos^2(x): Get Answer Here

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SUMMARY

The integration of the function f(x)=tan^5(x)/cos^2(x) results in tan^6(x)/6 + C. The substitution u = tan x simplifies the integral, where du = sec^2 x dx = 1/cos^2 x dx. Additionally, for the integral of ln(x^2-1), the method of polynomial long division and partial fractions is recommended, leading to the expression ln(x^2-1) = ln(x+1) + ln(x-1).

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Yankel
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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...
 
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If you make the substitution $u = \tan x$, then $ \displaystyle du = \sec^{2} x \ dx = \frac{1}{\cos^{2} x} \ dx $.
 
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 !
 
Using polynomial long division and then the method of partial fractions,

$$\frac{x^{2}}{x^{2}-1} = 1 + \frac{1}{x^{2}-1} = 1 + \frac{1}{2} \Big( \frac{1}{x-1} - \frac{1}{x+1} \Big)$$
 
Yankel said:
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!
 
Yankel said:
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 !
$$\ln(x^2-1)=\ln(x+1)+\ln(x-1).$$
 

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