`int t sec^2 2t dt` Evaluate the integral

`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`

`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`