# Math Help - Having trouble simplifying this integral

1. ## Having trouble simplifying this integral

So I have to evluate the integral and simplify it. I'm stuck near the end after applying F(b) - F(a) and can't simplify it further.

2. $\displaystyle \int (y^2-\sin y)~dy =\frac{y^3}{3}+\cos y \neq \frac{y^3}{3}-\cos y$

3. Wait so it's + cos y and not - cos y? But to get -sin y it has to be -cos y.

4. Originally Posted by solidstatemath
Wait so it's + cos y and not - cos y? But to get -sin y it has to be -cos y.
I understand it as

$\displaystyle \frac{d}{dy}\sin y =\cos y$

and

$\displaystyle \frac{d}{dy}\cos y =-\sin y$

therefore

$\displaystyle \int-\sin y ~dy = \cos y +C$

5. I'm having trouble simplifying this.

I got:

[(4pi/5)^3 / 3] - [COSpi4/5] - [(-4pi/5)^3 / 3] - [-COS(-4pi/5)]

I think the COS term will cancel out and then we get 2[(4pi/5)^3 / 3] as the answer.

Things I am not sure about: the signs. I'm not sure about:

- [-COS(-4pi/5)] *** does this become positive? What about the -4pi/5?

6. Okay I think I got it, the COS part cancel out leaving just the 4pi/5 parts.

My end result is then:

[(4pi/5)^3 / 3] + [(4pi/5)^3 / 3]

I am trying to simplify this and have some trouble. I turned it into:

2 * [(4pi/5^3) * 3]

Can it be simplified further?

7. Originally Posted by solidstatemath
I'm having trouble simplifying this.

I got:

[(4pi/5)^3 / 3] - [COSpi4/5] - [(-4pi/5)^3 / 3] - [-COS(-4pi/5)]

I think the COS term will cancel out and then we get 2[(4pi/5)^3 / 3] as the answer.

Things I am not sure about: the signs. I'm not sure about:

- [-COS(-4pi/5)] *** does this become positive? What about the -4pi/5?
cosine is an even function ... $\cos(-x) = \cos(x)$

8. Skipping the details here, the format above would then be the first term with pi - cosine term + second pi term + second cosine term.

The is the most I can simplify it: 2 * [(4pi/5^3) * 3]

How does this look? Basically the cosine terms cancel out and leaves just 2 times the pi terms.