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

Question

62032 views around the world

You can reuse this answer
Creative Commons License

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#