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How do I calculate the following integral:

$$\int^{{\pi/2}}_0\sin(x)\ \cos(x)\ \mathrm dx$$

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up vote 1 down vote accepted

Try $u(x)=\sin(x)$; then $u'(x) = \cos x$, and using the change of variables the integral can be written as $\int_{u(0)}^{u(\frac{\pi}{2})} u(x) u'(x) dx$. Now note that $ \frac{d}{dx} u^2(x) = 2 u(x) u'(x)$.

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HINT: $\sin{2x}=2\sin{x}\cos{x}$

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Another way would be considering $(\sin x)'=\cos x$, and so $$\int_0^\frac{\pi}{2}\sin xd(\sin x)=\int_0^1udu.$$

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