Integrating Secant Cubed Times Tangent Cubed

[apiwowar]

integral of sec³xtan³xdx

i tried to start this by rewriting secx as 1/cosx and that got me the integral of sin²x/cos⁶xdx

after that i tried a regular substitution but that didnt work since it doesn't take care of the sin³x on top

kinda stuck on where to go
any help?

 

[gabbagabbahey]

Hint: \sin^3(x)=\sin(x)[1-\cos^2(x)]

 

[physicsman2]

I would try: sec²x(sec x)(tan²x)(tan x)

 

[apiwowar]

first guys idea worked, thanks

 

[thrill3rnit3]

tan²x = sec²x-1

 

[physicsman2]

first guys idea worked

I know, just showing you another way of going about it, with thrill3rnit3 giving you a huge hint on how to go about it.