Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find, correct to one decimal place, the value of

$$\int_{0}^{60} 2\sin(x/2) \, dx.$$

Can someone please show me how this question is done. It would be very helpful thanks!

share|cite|improve this question
To re-confirm, the ranges are in radian, right? – lab bhattacharjee Aug 6 '13 at 13:40
What do you know about integration? – Daniel Littlewood Aug 6 '13 at 13:42
Yes the original bound were in radians. But i converted it – Red Queen10101 Aug 6 '13 at 23:58
Sorry for the late reply, totally forgot about this. – Red Queen10101 Aug 7 '13 at 0:02
up vote 2 down vote accepted

To solve this integral one must use a substitution. If we set $u=x/2$, then $du=(1/2)dx$, which gives $2\,du=dx$. Thus, the integral then becomes, $$2\int_0^{30} 2\sin(u)\,du.$$ To get the limits on the integral, we used our substitution again, that is, at $x=0$, $u=0/2=0$ and at $x=60$, $u=60/2=30$. With all the pieces determined, the integral evaluates to,


share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.