Calculus: Early Transcendentals, Chapter 7, 7.2, Section 7.2, Problem 13

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