∫(\sin x)²⋅dx

$I:=∫(\sin x)²⋅dx=?$

It is possible to solve using partial integration directly.

However, I find it easier to first replace the sin²:

$(\cos x)²+(\sin x)²=1$ normation.

$(\cos x)²-(\sin x)²=\cos(x+x)$ cosine addition theorem.

Hence, it follows that:

$-(\sin x)²=\cos(2⋅x)-(\cos x)²$

$(\sin x)²=(\cos x)²-(\cos(2⋅x))$

$(\sin x)²=1-(\sin x)²-(\cos(2⋅x))$

$2⋅(\sin x)²=1-(\cos(2⋅x))$

$(\sin x)²=÷{1-(\cos(2⋅x))}{2}$

And thus:

$∫(\sin x)²⋅dx=∫÷{1-(\cos(2⋅x))}{2}⋅dx$

$I=÷{1}{2}⋅∫(1-(\cos(2⋅x)))⋅dx$

$I=÷{1}{2}⋅(∫1⋅dx-∫\cos(2⋅x)⋅dx)$

$I=÷{1}{2}⋅∫1⋅dx-÷{1}{2}⋅∫\cos(2⋅x)⋅dx$

$I=÷{1}{2}⋅x-÷{1}{2}⋅÷{\sin(2⋅x)}{2}$

Author: Danny (remove the ".nospam" to send)

Last modification on: Thu, 09 May 2013 .