I have been trying to solve the following problem all day?
$$∫\frac{(9x^4\cdot\sin(9x))}{(1+x^8)}dx$$
I made
$$u=\sin(9x)$$
but I'm not really getting anywhere with that. Any kind of feedback or guidance would be very appreciated!
EDIT: the interval is [-pi/2, pi/2]
This was written before the edit told that the integral has bounds.
This is a true monster, for sure.
What you could do is to use partial fraction decomposition to write $$\frac{9x^4}{1+x^8}=\sum_{i=1}^8 \frac{A_i}{x-r_i}$$ (have fun !) and then the problem reduces to $$\int\frac{(9x^4\cdot\sin(9x))}{(1+x^8)}\,dx=\sum_{i=1}^8 A_i\int \frac{\sin(9x)}{x-r_i}\,dx=\sum_{i=1}^8 A_i I_i$$ Now, consider $$I_i=\int \frac{\sin(9x)}{x-r_i}\,dx$$ and change variable $x=y+r_i$ to get $$I_i=\int \frac{\sin(9(y+r_i))}{y}\,dy=\cos(9r_i)\int \frac{\sin(9y)}{y}\,dy+\sin(9r_i)\int \frac{\cos(9y)}{y}\,dy$$ Using now $y=\frac z9$,we get $$I_i=\cos(9r_i)\int \frac{\sin(z)}{z}\,dz+\sin(9r_i)\int \frac{\cos(z)}{z}\,dz$$ that is to say $$I_i=\cos(9r_i)\,\text{Si}(z)+\sin(9r_i)\,\text{Ci}(z)$$ where appear the sine and cosine integrals.
Then, the real mess given by Wolfram Alpha as early commented by Ross Millikan.