One helpful substitution is to note that sin^2(x) = (1-cos(2x))/2
That makes the problem
∫x(1-cos(2x))/2 dx = 1/2 ∫ (x - x cos2x) dx = 1/2 ∫x dx - 1/2 ∫x cos2x dx
The firrst part is trivial, and the second can be done using integration by parts. Let
u = x
dx
dv = cos2x dx
v = 1/2 sin2x
Then you have
∫ x cos 2x dx = 1/2 x sin2x - ∫1/2 sin2x dx
= 1/2 x sin2x + 1/4 cos2x
Now just put it all together.
integral (xsin^2(x))dx
2 answers
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Remark:
sin ( x ) sin ( y ) = ( 1 / 2 ) [ cos ( y − x ) − cos ( y + x ) ]
sin² ( x ) = sin ( x ) ∙ sin ( x ) = ( 1 / 2 ) [ cos ( x − x ) − cos ( x + x ) ] =
( 1 / 2 ) [ cos ( 0 ) − cos ( 2x ) ] =
( 1 / 2 ) [ 1 − cos ( 2 x ) ]
So:
sin² ( x ) = ( 1 / 2 ) [ 1 − cos ( 2 x ) ]
Integration online
When you see list of results click on:
Integral Calculator • With Steps!
When page be open in rectangle type:
(xsin^2(x))
and click option: Go
When you see result clik: Show steps
Remark:
sin ( x ) sin ( y ) = ( 1 / 2 ) [ cos ( y − x ) − cos ( y + x ) ]
sin² ( x ) = sin ( x ) ∙ sin ( x ) = ( 1 / 2 ) [ cos ( x − x ) − cos ( x + x ) ] =
( 1 / 2 ) [ cos ( 0 ) − cos ( 2x ) ] =
( 1 / 2 ) [ 1 − cos ( 2 x ) ]
So:
sin² ( x ) = ( 1 / 2 ) [ 1 − cos ( 2 x ) ]