Integration by u substitution

[erpanicco]

This problem was in the integration by substitution set in my book, but I'm unsure how to set it up

(x^2 * sinx) / (1 + x^6)

I started by trying to make u=x^3 which gave

1/3 sinx/(1 + u^2) du but then I'm still left with an x. Is there a better value for u?

Thanks

 

[galactus]

$\displaystyle \int\frac{x^{2}sin(x)}{x^{6}+1}dx$

This one is a booger and not easily doable by elementary means. This was in a regular calc book under u subs?.

Go here and enter in x^2*Sin[x]/(x^6+1) and you'll see what I mean.

http://integrals.wolfram.com/index.jsp

No limits of integration were given?.

 

[erpanicco]

Alright. Thanks for the heads up. I was able to do the rest of the u substitutions so I guess I'll just forget this one and move on

 

[galactus]

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.