$\displaystyle \int 16sin^2 x cos^2 x dx$ Would someone like to help a confused AB Calc Student?
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Originally Posted by penguinpwn $\displaystyle \int 16sin^2 x cos^2 x dx$ Would someone like to help a confused AB Calc Student? $\displaystyle \sin^2(x)\cos^2(x)=\left(\sin(x)\cos(x)\right)^2=\ left(\frac{1}{2}\sin(2x)\right)^2=\frac{1}{4}\sin^ 2(2x)=\frac{1-\cos(4x)}{8}$
Could you please explain where the third step you have shown came from?
Originally Posted by penguinpwn Could you please explain where the third step you have shown came from? $\displaystyle \sin(2x)=2\sin(x)\cos(x)\implies \sin(x)\cos(x)=\frac{1}{2}\sin(2x)$
Ok, i get it now the double angle formula. Thanks for clarifying for me =)
Originally Posted by penguinpwn Ok, i get it now the double angle formula. Thanks for clarifying for me =)
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