INTEGRATE THIS......

Manamy...actually Latex is not working..so i am unable to post the solution..btw i can tell u the method how to solve this question..
break ∫e^-x(sinx)ⁿ = ∫e^-x(sinx)^n-1 * sinx

= e^-x(sinx)^n-1 ∫ sinxdx - ∫[e^-x(sinx)^n-2 (n-1)cosx - e^-x(sinx)^n-1]cosx

= -e^-x(sinx)^n-1cosx - ∫[e^-x(sinx)^n-2 (n-1)cos²x - e^-x(sinx)^n-1cosx]

write cos²x = 1 - sin²x and expand this integral ∫e^-x(sinx)^n-1cosx]using by parts..take cosx as second function..
At last u will get the relation
I_n + (n-1)I_n + I_n/n = (n-1)I_n-2
put n = 10 and u will get the relation 101I₁₀/I₈ = 90