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#