Evaluate the integral $int \ sinx/(2+cos^2x) \ dx$ ?

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Answer 1 · 2018-01-14T01:21:44

$int \ sinx/(2+cos^2x) \ dx = -sqrt(2)/2 arctan(cosx/sqrt(2)) + C$

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:

$I = -sqrt(2)/2 arctan(cosx/sqrt(2)) + C$

Evaluate the integral int \ sinx/(2+cos^2x) \ dx int \ sinx/(2+cos^2x) \ dx ?