i need help with this integral....i seem to be getting no where with it (1/8)[integral]sin^2(2x)cos(2x) thanks
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Originally Posted by johntuan i need help with this integral....i seem to be getting no where with it (1/8)[integral]sin^2(2x)cos(2x) thanks substitute $\displaystyle \sin 2x = t$ so that, $\displaystyle 2\cos 2x \;dx = dt $ $\displaystyle \cos 2x \;dx = \frac{dt}{2} $ Spoiler: $\displaystyle \frac{1}{8}\int \sin ^2 2x\; \cos 2x~dx=\frac{1}{16}\int t^2~dt$
Originally Posted by johntuan i need help with this integral....i seem to be getting no where with it (1/8)[integral]sin^2(2x)cos(2x) thanks $\displaystyle sin^2(2x) = 1-cos^2(2x)$ $\displaystyle \frac{1}{8} \int cos(2x) - cos^3(2x)$ Which should be easier to integrate
Originally Posted by e^(i*pi) $\displaystyle sin^2(2x) = 1-cos^2(2x)$ $\displaystyle \frac{1}{8} \int cos(2x) - cos^3(2x)$ Which should be easier to integrate if i continued to integrate would it lead me to the answer of (1/8)[(1/3)(1/2)sin^3(2t)]......? thats the answer...I'm not sure...
Originally Posted by johntuan if i continued to integrate would it lead me to the answer of (1/8)[(1/3)(1/2)sin^3(2t)]......? thats the answer...I'm not sure... Ignore my answer lol, the guy's above me is much simpler XD
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