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

  1. #1
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    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?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by penguinpwn View Post
    \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}
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  3. #3
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    Could you please explain where the third step you have shown came from?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by penguinpwn View Post
    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)
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  5. #5
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    Ok, i get it now the double angle formula. Thanks for clarifying for me =)
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by penguinpwn View Post
    Ok, i get it now the double angle formula. Thanks for clarifying for me =)
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