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find the antiderivitive of sin^3(x)
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Use the identity \[\cos^2(x) + \sin^2(x) = 1\] so \[\int\limits_{}^{} \sin^3(x) dx = \int\limits_{}^{} (1-\cos^2(x)) \sin(x) dx\] try doing that integral now
Cheers thanks, I'll let you know
your welcome
I'm really stumped, can you help me out some more?
use a u-substitution u = 1 - cos^2(x) du = -2cos(x) -du/2 = cos(x) try doing it now
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Thanks
Do I change cos^2 (x) into 1/2[1+cos(2x)]?
actually try u= cos x then du = - sin x dx and then you integral becomz \(\large - \int (1-u^2)du\) which is pretty easy to integrate :)
Thankyou :)
welcome ^_^ let us know if you get stuck in any step ...
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