Evaluate the integral #int \ sinx/(2+cos^2x) \ dx #?
1 Answer
Jan 14, 2018
# int \ sinx/(2+cos^2x) \ dx = -sqrt(2)/2 arctan(cosx/sqrt(2)) + C #
Explanation:
We seek:
# I = int \ sinx/(2+cos^2x) \ dx #
If we look at the denominator then a substitution:
# 2u^2 = cos^2x #
looks promising, so let us try the substitution:
# u = cosx/sqrt(2) => 2u^2 = cos^2x #
Differentiating implicitly wrt
# (du)/dx = -sinx/sqrt(2) #
So substituting into the integral, it becomes:
# I = int \ (-sqrt(2))/(2+2u^2) \ du #
# \ \ = -sqrt(2) \ int \ 1/(2(1+u^2)) \ du #
# \ \ = -sqrt(2)/2 \ int \ 1/(1+u^2) \ du #
This is now a standard result, thus:
# I = -sqrt(2)/2 arctan(u) + C #
Then restoring the substitution:
# I = -sqrt(2)/2 arctan(cosx/sqrt(2)) + C #