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

1. ## 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?

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

Would someone like to help a confused AB Calc Student?
$\sin^2(x)\cos^2(x)=\left(\sin(x)\cos(x)\right)^2=\ left(\frac{1}{2}\sin(2x)\right)^2=\frac{1}{4}\sin^ 2(2x)=\frac{1-\cos(4x)}{8}$

3. Could you please explain where the third step you have shown came from?

4. Originally Posted by penguinpwn
Could you please explain where the third step you have shown came from?
$\sin(2x)=2\sin(x)\cos(x)\implies \sin(x)\cos(x)=\frac{1}{2}\sin(2x)$

5. Ok, i get it now the double angle formula. Thanks for clarifying for me =)

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