How do I solve` int_0^((3pi)/2)|cosx|dx` ?



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Posted on (Answer #1)

We can solve this by breaking into two cases depending on whether  `cosx` is positive or negative.

We know that `cosx` for `x in [0,pi/2)` and negative for`x in(pi/2,(3pi)/2)`, so for  `x in [0,pi/2)` `|cosx|=cosx` and for  `x in(pi/2,(3pi)/2)` `|cosx|=-cosx`, thus we have:



Note: Above we have used the following property of integral:

Let `f` be integrable on segment `[a,b]` and let `c in (a,b)` then we have


For more on definite integrals see the link below.

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