Integration involving trig functions and various powers of X

[seanoe25]

∫[6x^6 sin (9x)]/[1+x^10] * dx

I've set u =x^6
du=6x^5*dx
dx=du/6x^5

∫[6x^6 sin (9x)]/[1+x^10] * (du/6x^5)
=
∫[x*sin(9x)*du]/1+x^10.

Can someone help me figure out the next step? I'm thinking of putting a constant out in front, so I can use 2du for (x^10)

 

[SammyS]

∫[6x^6 sin (9x)]/[1+x^10] * dx

I've set u =x^6
du=6x^5*dx
dx=du/6x^5

∫[6x^6 sin (9x)]/[1+x^10] * (du/6x^5)
=
∫[x*sin(9x)*du]/1+x^10.

Can someone help me figure out the next step? I'm thinking of putting a constant out in front, so I can use 2du for (x^10)

I doubt that this will work. That's a messy integral.

However, you haven't finished the substitution ! You should end up with an integral in the variable, u, with no x what-so-ever.

Are you sure you have written the problem correctly?

What topics are you currently covering in whatever class this problem is from?

 

[seanoe25]

Yeah, sadly I have written it down properly. Right now we're covering how to solve definite integrals with the use of substitution; it's a beautiful thing when it works, but these problems are moral-breakers. It's acually:

∫ 6x^6 sin(9x)/[1+x^10] *dx

with the upper limit set at pi/2, and the lower limit at -pi/2.

I excluded the limits part because I felt once I got help with the substitution, I had the problem down. But finding the right u is very difficult.

 

[SammyS]

Yeah, sadly I have written it down properly. Right now we're covering how to solve definite integrals with the use of substitution; it's a beautiful thing when it works, but these problems are moral-breakers. It's acually:

∫ 6x^6 sin(9x)/[1+x^10] *dx

with the upper limit set at pi/2, and the lower limit at -pi/2.

I excluded the limits part because I felt once I got help with the substitution, I had the problem down. But finding the right u is very difficult.

Having it be a definite integral makes all the difference in the world!

Is the integrand either an even or an odd function?

 

[seanoe25]

A glimpse of hope! I believe the integrand is an even function. Because when I plugged in f(-x), everything came out to be the same

 

[SammyS]

A glimpse of hope! I believe the integrand is an even function. Because when I plugged in f(-x), everything came out to be the same

Not quite!

sin(-9x) = -sin(9x)

Graph the integrand.