trig integral

**johntuan** · April 11th 2009, 06:22 PM

i need help with this integral....i seem to be getting no where with it

(1/8)[integral]sin^2(2x)cos(2x)

thanks

**Shyam** · April 11th 2009, 06:33 PM

Originally Posted by johntuan

i need help with this integral....i seem to be getting no where with it

(1/8)[integral]sin^2(2x)cos(2x)

thanks

substitute

$\sin 2x = t$

so that,

$<br /> 2\cos 2x \;dx = dt<br />$

$<br /> \cos 2x \;dx = \frac{dt}{2}<br />$

Spoiler:

**e^(i*pi)** · April 11th 2009, 06:34 PM

Originally Posted by johntuan

i need help with this integral....i seem to be getting no where with it

(1/8)[integral]sin^2(2x)cos(2x)

thanks

$sin^2(2x) = 1-cos^2(2x)$

$\frac{1}{8} \int cos(2x) - cos^3(2x)$

Which should be easier to integrate

**johntuan** · April 11th 2009, 06:44 PM

Originally Posted by e^(i*pi)

$sin^2(2x) = 1-cos^2(2x)$

$\frac{1}{8} \int cos(2x) - cos^3(2x)$

Which should be easier to integrate

if i continued to integrate would it lead me to the answer of

(1/8)[(1/3)(1/2)sin^3(2t)]......?

thats the answer...I'm not sure...

**e^(i*pi)** · April 11th 2009, 06:48 PM

Originally Posted by johntuan

if i continued to integrate would it lead me to the answer of

(1/8)[(1/3)(1/2)sin^3(2t)]......?

thats the answer...I'm not sure...

Ignore my answer lol, the guy's above me is much simpler XD

Thread: trig integral