`int_0^((3pi)/2) |sin(x)| dx` Evaluate the integral

Expert Answers
tiburtius eNotes educator| Certified Educator

We have to be careful here because of absolute value sign. We need to know where the function under absolute value changes sign (in the interval of integration), which is at `x=pi` in this case. Therefore, we have

`int_0^((3pi)/2)|sin(x)|dx=int_0^pi sin(x)dx+int_pi^((3pi)/2)(-sin(x))dx=`

`-cos(x)|_0^pi+cos(x)|_pi^((3pi)/2)=-cos(pi)+cos(0)+cos((3pi)/2)-cos(pi)=`

`-(-1)+1+0-(-1)=3` <-- The solution