How do you evaluate $intcos(x)/(9+sin^2x)dx$ ?

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Answer 1 · 2017-03-06T22:17:44

$1/3tan^-1(sinx/3)+C$

$intcosx/(9+sin^2x)dx$

Let $u=sinx$ so $du=cosxcolor(white).dx$ :

$=int1/(9+u^2)du=1/9int1/(1+u^2/9)du=1/9int1/(1+(u/3)^2)du$

Now let $v=u/3$ so $dv=1/3du$ :

$=1/3int(1/3)/(1+(u/3)^2)du=1/3int1/(1+v^2)dv$

Which is a common integral:

$=1/3tan^-1(v)=1/3tan^-1(u/3)=1/3tan^-1(sinx/3)+C$

How do you evaluate intcos(x)/(9+sin^2x)dxintcos(x)/(9+sin^2x)dx?