How do you find the integral of ${sin}^{2} (3 x) d x$ $sin^2(3x)dx$ ?

Question

63107 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-12T22:02:34

$= \frac{1}{2} x - \frac{1}{12} sin 6 x + C$ $=1/2x - 1/12sin6x + C$

$int \ sin^2 (3x) \ dx$

small book keeping gesture is to make the sub $u = 3x, du = 3 dx$

$1/3int \ sin^2 (u) \ du$

then we use the cosine double angle formulae

$cos 2A = 1 - 2 sin^2 A$

so $sin^2 A = (1 - cos 2A)/2$

$=1/6int \ 1 - cos 2u \ du$

$=1/6( u - 1/2sin 2u ) + C$

$=1/6( 3x - 1/2sin( 2*3x) ) + C$

$=1/2x - 1/12sin6x + C$

How do you find the integral of sin^2(3x)dxsin2(3x)dxsin^2(3x)dx?