O lvl A Maths: Integration

Question from http://sgforums.com/forums/2297/topics/345581

Find d/dx (sin³ 2x) and hence evaluate $\int_{0}^{\pi /4}{\cos^{3} 2x }dx$ .

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Answer:

d/dx (sin³ 2x)
= (3sin² 2x)(2cos 2x)
= 6(sin² 2x)(cos 2x)

Hence,

$\int_{0}^{\pi /4}{\cos^{3} 2x }dx$
$=\int_{0}^{\pi /4}{\cos 2x \left (\cos^{2} 2x \right )}dx$
$=\int_{0}^{\pi /4}{\cos 2x \left (1-\sin^{2} 2x \right )}dx$
$=\int_{0}^{\pi /4}{\cos 2x -\cos 2x\sin^{2} 2x }dx$
$=\int_{0}^{\pi /4}{\cos 2x }dx-\int_{0}^{\pi /4}{\cos 2x \sin^{2}2x}dx$
$=\left[\frac{1}{2}\sin 2x \right]_{0}^{\frac{\pi }{4}}-\frac{1}{6}\int_{0}^{\frac{\pi }{4}}6\cos 2x\sin ^{2}2x dx$
$=\left[\frac{1}{2}\sin 2x \right]_{0}^{\frac{\pi }{4}}-\left[\frac{1}{6}\sin ^{3}2x dx \right]_{0}^{\frac{\pi }{4}}$