Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 13

inttsec^2(2t)dt
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
Now using the above method of integration by parts,
inttsec^2(2t)dt=tintsec^2(2t)dt-int(d/dt(t)intsect^2(2t))dt
=t*tan(2t)/2-int(1*tan(2t)/2)dt
=1/2t*tan(2t)-1/2inttan(2t)dt
Now let's evaluate inttan(2t)dt by using the method of substitution,
Substitute x=cos(2t),=>dx=-2sin(2t)dt
inttan(2t)dt=int(sin(2t)/cos(2t))dt
=intdx/(-2x)
=-1/2ln|x|
substitute back x=cos(2t)
=-1/2ln|cos(2t)|
inttsec^2(2t)dt=1/2t*tan(2t)-1/2(-1/2ln|cos(2t)|+C
C is a constant
inttsec^2(2t)dt=1/2t*tan(2t)+1/4ln|cos(2t)|+C