Tuesday, 10 May 2016

Grove HSC 3U - Challenge 5, Q23

hey sir also its in challenge 5



$u=\cos x$
$du=-\sin x dx$

Recall $\sin^2x=1-\cos^2x$

So, $I=-\int \sin^2x\, \sin x \,dx$
$I=-\int (1-\cos^2x)\, \sin x \,dx$
$I=-\int (1-u^2)\, du$
$I=-u+u^3/3+c$

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