What is $inttan(x)^3sec(x)^3dx$ ?

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Answer 1 · 2018-08-07T22:19:38

$inttan(x)^3sec(x)^3dx=1/5sec(x)^5-1/3sec(x)^3+C$

$I=inttan(x)^3sec(x)^3dx$

$=inttan(x)tan(x)^2sec(x)^3dx$

Because $tan(x)^2=sec(x)^2-1$

$I=inttan(x)sec(x)^3(sec(x)^2-1)dx$

$=inttan(x)sec(x)^5dx-inttan(x)sec(x)^3dx$

$=-int(-sin(x))/(cos(x)^6)dx+int(-sin(x))/(cos(x)^4)dx$

Let $u=cos(x)$
$du=-sin(x)dx$

So:

$I=int1/u^4du-int1/u^6du$

$=1/(5u^5)-1/(3u^3)+C$ , $C in RR$

$=1/(5cos(x)^5)-1/(3cos(x)^3)+C$ , $C in RR$

$=1/5sec(x)^5-1/3sec(x)^3+C$ , $C in RR$

What is inttan(x)^3sec(x)^3dxinttan(x)^3sec(x)^3dx ?