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

2 Answers
Feb 11, 2018

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

Feb 11, 2018

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