Question

What is the integration of `sec^2 x cos^2 (2x) dx`?

Answer 1

`I=sin(2x)-2x+tan(x)+C`

Explanation:

We want to solve

`I=intsec^2(x)cos^2(2x)dx`

Rewrite the integrand using `color(blue)(cos(2x)=2cos^2(x)-1)`

`I=intsec^2(x)(2cos^2(x)-1)^2dx`

`=intsec^2(x)(4cos^4(x)-4cos^2(x)+1)dx`

`=int4cos^2(x)-4+sec^2(x)dx`

Use the sum rule for integrals

`I=4intcos^2(x)dx-4intdx+intsec^2(x)dx`

Rewrite the integrand (1) using `color(blue)(cos(2x)=1/2(1+cos(2x)))`

`I=2int(1+cos(2x))dx-4intdx+intsec^2(x)dx`

`=2intcos(2x)dx-2intdx+intsec^2(x)dx`

Integrate (Notice `color(blue)((tan(x))'=sec^2(x))`)

`I=sin(2x)-2x+tan(x)+C`

Answer 2

`intsec^2xcos^2(2x)dx=-2x+sin2x+tanx`

Explanation:

Find:
`intsec^2xcos^2(2x)dx`

`sec^2x=1/cos^2x`

`cos^2(2x)=(cos2x)^2=(2cos^2x-1)^2`
Thus

`intsec^2xcos^2(2x)dx=int1/cos^2x(2cos^2x-1)^2dx`
`=int(2cos^2x-1)^2/cos^2xdx`
`=int(4cos^4x-4cos^2x+1)/cos^2xdx`
`=int(4cos^2x-4+sec^2x)dx`
`=int4cos^2xdx-int4dx+intsec^2xdx`
`int4cos^2xdx`
`cos^2x=1/2(1+cos2x)`
`4cos^2xdx=4(1/2(1+cos2x))=2+2cos2x`
`int4cos^2xdx=int(2+2cos2x)dx`
`=int2dx+int2cos2xdx`
Thus,
`int4cos^2xdx-int4dx+intsec^2xdx=`
`int2dx+int2cos2xdx-int4dx`
`+intsec^2xdx`
`=2x+sin2x-4x+tanx`
`=-2x+sin2x+tanx`

Now,

`intsec^2xcos^2(2x)dx=-2x+sin2x+tanx`