# Math Help - integral abs sinx [0,3pi/2]

1. ## integral abs sinx [0,3pi/2]

integral abs sinx [0,3pi/2]

would this be -cosx(abs sinx/sinx)?

2. Originally Posted by kensington
integral abs sinx [0,3pi/2]

would this be -cosx(abs sinx/sinx)?
Draw the graph, then set up and evaluate the required integrals.

3. You cannot ignore the absolute value.
Break up the integral into the parts where sin x is positive and negative and flip the negative part over.
Then you can integrate.

4. what do you mean flip the negative part over?

5. Originally Posted by kensington
what do you mean flip the negative part over?
This means that [LaTeX ERROR: Convert failed]

6. Originally Posted by Vlasev
This means that [LaTeX ERROR: Convert failed]
soo then is it

abs sinx= sinx if x=> 3pi/2
=-sinx if x< 3pi/2

so then integral from 0 to pi sinx+ integral from pi to 3pi/2 -sinx
which is then [-cosx]+ [cosx]
[(-cos0)-(-cospi)]+ [(cospi)-(cos3pi/2)]
[-1-(1)] + [-1-(-1/2)]
-2-1/2
=-2.5??

Would that be correct?

7. This may surprise you but: $\int_0^{\frac{{3\pi }}
{2}} {\left| {\sin (x)} \right|dx} = 3\int_0^{\frac{\pi }
{2}} {\sin (x)dx}$
.

Think about it after looking at the graph.

8. Originally Posted by Plato
This may surprise you but: $\int_0^{\frac{{3\pi }}
{2}} {\left| {\sin (x)} \right|dx} = 3\int_0^{\frac{\pi }
{2}} {\sin (x)dx}$
.

Think about it after looking at the graph.
so then would it be
3 [[(-cos0)-(-cospi/2)]+ [(coso)-(cospi/2)]]?

9. No. The answer is 3.

10. Originally Posted by Plato