1. ## definite integral

evaluate the definite integral:

I tried using substitution where U=X^8 and du=8xd^7 but I don't really know where to go with this problem

the integral between pi/2 to -pi/2
 x8 sin x 1 + x0
dx

(Sorry i couldn't figure out how to get the integral sign in)
thank you!

2. ## Re: definite integral

$\displaystyle I=\int\frac{x^8\sin(x)}{1+x^0}\,dx=\frac{1}{2}\int x^8\sin(x)\,dx$

then I would try integration by parts, and a bit of experience tells me we will need two formulas:

Let:

$\displaystyle I_1=\int x^n\sin(x)\,dx$

$\displaystyle u=x^n\:\therefore\:du=nx^{n-1}\,dx$

$\displaystyle dv=\sin(x)\,dx\:\therefore\:v=-\cos(x)$

Hence:

$\displaystyle I_1=-x^n\cos(x)+n\int x^{n-1}\cos(x)\,dx$

Now use integration by parts again to develop a formula for the resulting integral on the right:

$\displaystyle I_2=\int x^n\cos(x)\,dx$

$\displaystyle u=x^n\:\therefore\:du=nx^{n-1}\,dx$

$\displaystyle dv=\cos(x)\,dx\:\therefore\:v=\sin(x)$

Hence:

$\displaystyle I_2=x^n\sin(x)-n\int x^{n-1}\sin(x)\,dx$

Now apply these formulas alternately until $\displaystyle n-1=0$ and you may evaluate the integral of a trig function alone to get the final result.