How do you find the integral of $sin (2 x) {cos}^{2} (2 x) d x$ $sin(2x)cos^2(2x)dx$ ?

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Answer 1 · 2016-09-09T19:50:36

$\int sin (2 x) {cos}^{2} (2 x) d x$ $int sin(2x)cos^2(2x)dx$

$= - \frac{1}{2} \int - 2 sin (2 x) {cos}^{2} (2 x) d x$ $= -1/2int-2sin(2x)cos^2(2x)dx$

$u = cos (2 x)$ $u = cos(2x)$
$d u = - 2 sin (2 x) d x$ $du = -2sin(2x)dx$

Now we have

$- \frac{1}{2} \int u^{2} d u$ $-1/2intu^2du$

and as you know it's

$- \frac{1}{6} [u^{3}] + C$ $-1/6[u^3] + C$

when you substitute back you have

$- \frac{1}{6} [{cos}^{3} (2 x)] + C$ $-1/6[cos^3(2x)] + C$

How do you find the integral of sin(2x)cos2(2x)dxsin(2x)cos^2(2x)dx?