# One more for tonight, then I'll leave you guys alone...

• Mar 9th 2010, 03:57 PM
penguinpwn
One more for tonight, then I'll leave you guys alone...
$\int 16sin^2 x cos^2 x dx$

Would someone like to help a confused AB Calc Student?
• Mar 9th 2010, 04:01 PM
Drexel28
Quote:

Originally Posted by penguinpwn
$\int 16sin^2 x cos^2 x dx$

Would someone like to help a confused AB Calc Student?

$\sin^2(x)\cos^2(x)=\left(\sin(x)\cos(x)\right)^2=\ left(\frac{1}{2}\sin(2x)\right)^2=\frac{1}{4}\sin^ 2(2x)=\frac{1-\cos(4x)}{8}$
• Mar 9th 2010, 04:14 PM
penguinpwn
Could you please explain where the third step you have shown came from?
• Mar 9th 2010, 04:16 PM
Drexel28
Quote:

Originally Posted by penguinpwn
Could you please explain where the third step you have shown came from?

$\sin(2x)=2\sin(x)\cos(x)\implies \sin(x)\cos(x)=\frac{1}{2}\sin(2x)$
• Mar 9th 2010, 04:19 PM
penguinpwn
Ok, i get it now the double angle formula. Thanks for clarifying for me =)
• Mar 9th 2010, 04:22 PM
Drexel28
Quote:

Originally Posted by penguinpwn
Ok, i get it now the double angle formula. Thanks for clarifying for me =)

(Yes)(Handshake)