`int t sin^2(t) dt` Evaluate the integral

`inttsin^2(t)dt`

Let's use the method of integration by parts,

If f(x) and g(x) are differentiable functions, then

`intf(x)g'(x)dx=f(x)g(x)-intf'(x)g(x)dx`

If we write f(x)=u and g'(x)=v, then

`intuvdx=uintvdx-int(u'intvdx)dx`

Let's denote u=t and `v=sin^2(t)`

`u'=d/dt(t)=1`

as

`inttsin^2(t)dt=tintsin^2(t)dt-int(intsin^2(t)dt)dt`

Noe let's use the identity :`sin^2(x)=(1-cos(2x))/2`

`=tint(1-cos(2t))/2dt-int(int((1-cos(2t))/2)dt)dt`

`=t/2int(1-cos(2t)dt-int(1/2int(1-cos(2t))dt)dt`

`=t/2(t-sin(2t)/2)-1/2int(t-sin(2t)/2)dt`

`=t/2(t-sin(2t)/2)-1/2(t^2/2-(-cos(2t)/4))`

`=t^2/2-(tsin(2t))/4-t^2/4-cos(2t)/8`

`=t^2/4-(tsin(2t))/4-cos(2t)/8`

Add a constant C to the solution,

`=t^2/4-(tsin(2t))/4-cos(2t)/8+C`