Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do I calculate the following integral:

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

share|improve this question
add comment

3 Answers

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)$.

share|improve this answer
add comment

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

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.