How do you evaluate #intcos(x)/(9+sin^2x)dx#?
1 Answer
Mar 6, 2017
Explanation:
#intcosx/(9+sin^2x)dx#
Let
#=int1/(9+u^2)du=1/9int1/(1+u^2/9)du=1/9int1/(1+(u/3)^2)du#
Now let
#=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#