How do you find the integral of $sin^2 (ax)$ ?

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Answer 1 · 2018-03-22T20:56:11

$int sin^2(ax) dx= ( ax-sin(ax)cos(ax) )/(2a)+C$

Use the trigonometric identity:

$sin^2(ax) = (1-cos(2ax))/2$

So:

$int sin^2(ax) dx= int (1-cos(2ax))/2dx$

$int sin^2(ax) dx= 1/2 int dx -1/2 int cos(2ax)dx$

$int sin^2(ax) dx= x/2 -1/(4a) int cos(2ax)d(2ax)$

$int sin^2(ax) dx= x/2 -1/(4a)sin(2ax) +C$

$int sin^2(ax) dx= x/2 -1/(2a)sin(ax)cos(ax) +C$

$int sin^2(ax) dx= ( ax-sin(ax)cos(ax) )/(2a)+C$

How do you find the integral of sin^2 (ax) sin^2 (ax)?