What is $int sec(x)^3tan(x)dx$ ?

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Answer 1 · 2018-08-07T22:35:32

$intsec(x)^3tan(x)dx=1/3sec(x)^3+C$ , $C in RR$

$I=intsec(x)^3tan(x)dx$

$=-int(-sin(x))/(cos(x)^4)dx$

Let $u=cos(x)$

$du=-sin(x)dx$

So:

$I=-int1/u^4du$

$=1/(3u^3)$

$=1/(3cos(x)^3)+C$ , $C in RR$

$=1/3sec(x)^3+C$ , $C in RR$

\0/ Here's our answer !

What is int sec(x)^3tan(x)dxint sec(x)^3tan(x)dx ?