What is the integral of $x * cos^2 (x)$ ?

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Answer 1 · 2016-06-30T09:10:51

$= 1/4 x sin 2x + 1/8 cos 2x + x^2/4+ C$

$int dx qquad x * cos^2 (x)$

it will be easier first to use the double angle formula $cos 2A = 2 cos^2 A - 1$ or $cos^2 A = (cos 2A + 1)/2$

so we are looking at

$1/2 int dx qquad color{red}{x cos 2x} + x$

as it's a composite term, we should do the red bit using IBP ie

$int u v' = uv - int u' v$

here

$u = x, u' = 1$
$v' = cos 2x, v = 1/2 sin 2x$

so

$1/2 int dx qquad color{red}{x cos 2x} + x$

$= 1/2 { 1/2 x sin 2x - int dx qquad 1/2 sin 2x +int dx qquad x }$

$= 1/2 { 1/2 x sin 2x + 1/4 cos 2x + x^2/2 } + C$

$= 1/4 x sin 2x + 1/8 cos 2x + x^2/4+ C$

What is the integral of x * cos^2 (x) x * cos^2 (x)?