What's the integral of $int tan(x)^3 dx$ ?

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Answer 1 · 2016-08-08T15:46:13

$inttan^3(x)dx=tan^2(x)/2+ln(abs(cos(x)))+C$

Recall that the Pythagorean identity tells us that $tan^2(x)=sec^2(x)-1$ . Thus we can say that:

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

Now we can split up the integral:

$=inttan(x)sec^2(x)dx-inttan(x)dx$

For the first integral, let $u=tan(x)$ so $du=sec^2(x)dx$ .

$=intudu-inttan(x)dx$

$=u^2/2-inttan(x)dx$

$=tan^2(x)/2-intsin(x)/cos(x)dx$

Here, let $v=cos(x)$ so $dv=-sin(x)dx$ , so:

$=tan^2(x)/2+int(-sin(x))/cos(x)dx$

$=tan^2(x)/2+int(dv)/v$

$=tan^2(x)/2+ln(absv)+C$

$=tan^2(x)/2+ln(abs(cos(x)))+C$

What's the integral of int tan(x)^3 dxint tan(x)^3 dx?