Question

integration of sec^4(2x)dx

Answer 1

Use the trigonometric identities:

#sin^2 alpha = (1-cos2alpha)/2#

#cos^2 alpha = (1+cos2alpha)/2#

to have:

#sin^4(2x) = (sin^2(2x))^2 = (1-cos(4x))^2/4#

#sin^4(2x) = (1-2cos(4x) +cos^2(4x))/4#

#sin^4(2x) = 1/4-(cos(4x))/2 +1/8 +(cos(8x))/8#

#sin^4(2x) = 3/8-(cos(4x))/2 +(cos(8x))/8#

Using the linearity of the integral:

#int sin^4(2x) dx= 3/8int dx-1/2 int cos(4x)dx +1/8 intcos(8x)dx#

#int sin^4(2x) dx= (3x)/8-1/8 int cos(4x)d(4x) +1/64 intcos(8x)d(8x)#

#int sin^4(2x) dx= (3x)/8-(sin(4x))/8 + (sin(8x))/64+C#