- Thread starter
- #1

#### paulmdrdo

##### Active member

- May 13, 2013

- 386

∫cos

^{7}2xdx = ∫(cos

^{6}2x cos2x)dx

=∫(1-sin

^{2}2x)

^{3}cos2x

let u = sin2x;

du = cos2xdx

dx = (du/2cos2x)

∫(1-u

^{2})

^{3}cos2x*du/2cos2x

1/2∫(1-u

^{2})

^{3}*du

1/2∫(1-u

^{2})

^{2}*(1-u

^{2})du

1/2∫[(1-3u

^{2}-u

^{4}-u

^{6}]du

1/2∫du-3/2∫u

^{2}du-1/2∫u

^{4}du-1/2∫u

^{6}du

1/2(u)+1/2(u

^{3})-1/10(u

^{5})-1/14(u

^{7})+C

1/2(sin2x)+1/2(sin

^{3}2x)-1/10(sin

^{5}2x)-1/14(sin

^{7}2x)+C

are they correct?