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...
∫[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)
actually I got it. I forgot arctan (u) = 1/1+u^2. So it becomes
-7 ∫arctan (u)*du
=
-7 arctan (cos x) +C
Thanks for the push. And putting up with my crappy algebra
I just forgot to type it in.
so now I have
-7 ∫1/(1+u^2), but this is where it gets hazy for me.
would it be
-7* 1/(-2+1) *u^(-2+1)?
which would equal
-7*-1*u^-1, which gives me 7/cos(x), but that doesn't seem right.
Don't be sorry, you all are helping me
I took the integral identity ∫f(x)+g(x)=∫f(x)+∫g(x), and somehow came up with some bogus new rule.
I used u =cos (x)
du= - sin (x)*dx
dx= du/-sin (X)
∫ 7 sin (x)*dx/[1+u^2]
∫ [7 sin (x)* [du/-sin (x)]/ 1+u^2
=
∫-7du/[1+u^2]
=
7∫1/[1+u^2]...
1.) ∫[(7 sin (x))/[1+cos^2(x)]] * dx
2.) I'm looking at the trig identity sin^2 x+cos^2 x=1, and am wondering if I could use that in solving the problem. Or should I use u=sin x, then du= cos x, then plug those in?
3.) so I thought maybe it would be easier to separate the two...