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 #x#:

# (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 #