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Ok, I know that sin 2x can be substituted out for 2 sin x cos x.
so now I have the Integral of (sin x ) ( 2 sin x cos x ) dx which is 2 sin2x cos x dx
If I use integration by parts with the u and dv, I find myself right back again..can you help. The book gives an answer, but I am not sure how it was achieved.
the book gives 2/3 sin 2x cos x - 1/3 cos 2x sin x + C I may be making this more difficult than it is ?
when you get the integral of 2 sin2x cos x dx use u substitution. u= sinx du= cosxdx. Then you'll get the integral of 2u^2 du.. .and then integrate...
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Integral of tan square x secant x?
convert tan^2x into sin^2x/cos^2x and secant x into 1/cos x combine terms for integral sin^2x/cos^3x dx then sub in u= cos^3x and du=-2sin^2x dx
Integral of sec2x-cosx plus x2dx?
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Integral of sin squared x?
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Find the derivative of y x2 sin x 2xcos x - 2sin x?
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Integration of sin2x?
Integral( sin(2x)dx) = -(cos(2x)/2) + C
