Evaluating the integral of x²sin(x)/(1+x⁶) from -π/2 to π/2

[sapiental]

the definite integral:

from a = (-pi/2) to (pi/2) f(((x^2)(sinx))/(1+x^6))dx

this is the way it seems most logic to me to set it up using substitution:

u = x
du = dx

from a = (-pi/2) to (pi/2) f(((u^2)(sinu))/(1+u^6))du

= (((-cos(u))(1/3u^3))/(u+1/7u^7))+CI know how to evaluate it from here, I just need some feedback on my substitution setup.

Thanks in advance.

 

[neutrino]

u = x
du = dx

That sort of substitution does nothing but change the letter denoting the variable.

 

[shmoe]

Try sketching the graph. Notice anything?

Take the square root of the numerator and denominator and use the substitution $$x^{3} = \tan \theta, x = \sqrt[3]{\tan \theta}$$ You end up getting $$\frac{1}{3} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \tan \theta$$

I'm worried about what "Take the square root of the numerator and denominator" could possibly mean? I don't see how your substitution gives what you claim it gives either.

 

[courtrigrad]

yeah, I made a mistake. I took the integral of the square root of the function instead of the actual function.