Actually, I have seen this integral before. The limits of integration were 0 to Pi/2. I bet that is what yours was and you left out the limits of integration. Am I right?.
The trick is observing that this makes it an odd function and it evaluates to 0.
But, if you would like to see what the indefinite form is. Take a gander at this monstrisity:
\(\displaystyle \frac{1}{6 \left(i+\sqrt{3}\right)}\left(\left(1-i \sqrt{3}\right) \text{CosIntegral}\left[-(-1)^{1/6}+x\right] \text{Sin}\left[(-1)^{1/6}\right]\)\(\displaystyle +\left(1-i \sqrt{3}\right) \text{CosIntegral}\left[(-1)^{1/6}+x\right] \text{Sin}\left[(-1)^{1/6}\right]-2 \left((-1)^{2/3} \text{CosIntegral}\left[-(-1)^{5/6}+x\right] \text{Sin}\left[(-1)^{5/6}\right]+(-1)^{2/3} \text{CosIntegral}\left[(-1)^{5/6}+x\right]\)\(\displaystyle \text{Sin}\left[(-1)^{5/6}\right]-i \text{CosIntegral}[-i+x] \text{Sinh}[1]+2 (-1)^{1/6} \text{CosIntegral}[-i+x] \text{Sinh}[1]+(-1)^{5/6}\)\(\displaystyle \text{CosIntegral}[-i+x] \text{Sinh}[1]-i \text{CosIntegral}[i+x] \text{Sinh}[1]+2 (-1)^{1/6} \text{CosIntegral}[i+x]\)\(\displaystyle \text{Sinh}[1]+(-1)^{5/6} \text{CosIntegral}[i+x] \text{Sinh}[1]+\text{Cosh}[1] \text{SinIntegral}[i-x]-(-1)^{1/3} \text{Cosh}[1]\)\(\displaystyle \text{SinIntegral}[i-x]+2 (-1)^{2/3} \text{Cosh}[1] \text{SinIntegral}[i-x]+2 \text{Cos}\left[(-1)^{1/6}\right] \text{SinIntegral}\left[(-1)^{1/6}-x\right]\)\(\displaystyle -2 (-1)^{1/3} \text{Cos}\left[(-1)^{1/6}\right] \text{SinIntegral}\left[(-1)^{1/6}-x\right]+(-1)^{2/3} \text{Cos}\left[(-1)^{1/6}\right]\)\(\displaystyle \text{SinIntegral}\left[(-1)^{1/6}-x\right]-(-1)^{2/3} \text{Cos}\left[(-1)^{5/6}\right] \text{SinIntegral}\left[(-1)^{5/6}-x\right]\)\(\displaystyle +\text{Cosh}[1] \text{SinIntegral}[i+x]-(-1)^{1/3} \text{Cosh}[1] \text{SinIntegral}[i+x]+2 (-1)^{2/3} \text{Cosh}[1] \text{SinIntegral}[i+x]\)\(\displaystyle +2 \text{Cos}\left[(-1)^{1/6}\right] \text{SinIntegral}\left[(-1)^{1/6}+x\right]-2 (-1)^{1/3} \text{Cos}\left[(-1)^{1/6}\right]\)\(\displaystyle \text{SinIntegral}\left[(-1)^{1/6}+x\right]+(-1)^{2/3} \text{Cos}\left[(-1)^{1/6}\right]\)\(\displaystyle \text{SinIntegral}\left[(-1)^{1/6}+x\right]-(-1)^{2/3} \text{Cos}\left[(-1)^{5/6}\right] \text{SinIntegral}\left[(-1)^{5/6}+x\right]\right)\right)\)
I did not type this all out. Mathematica allows one to copy in LaTex.
Now do you see why this problem was more than likely as I described?. The given integration limits were 0 to Pi/2. Seeing this shows it is an odd function and results to 0. Here is the graph. You can see how it is odd.
