What is the integral of #int tan^2(x)sec^4(x)#?

Question

3647 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-11T21:07:28

#inttan^2(x)sec^4(x)=tan^5(x)/5+tan^3(x)/3+C#

Recall that through the Pythagorean identity we know that #tan^2(x)+1=sec^2(x)#. Thus:

#I=inttan^2(x)sec^4(x)dx=inttan^2(x)sec^2(x)sec^2(x)dx#

Rewriting one:

#I=inttan^2(x)(tan^2(x)+1)sec^2(x)dx#

Keeping the other #sec^2(x)# as the derivative of #tan(x)#:

#I=int(tan^4(x)+tan^2(x))sec^2(x)dx#

Now, let #u=tan(x)# so #du=sec^2(x)dx#:

#I=int(u^4+u^2)du#

#I=u^5/5+u^3/3+C#

#I=tan^5(x)/5+tan^3(x)/3+C#