Integration- (Substitution Rule) Help

  • Thread starter Thread starter jordan123
  • Start date Start date
  • Tags Tags
    Integration
jordan123
Messages
15
Reaction score
0

Homework Statement


Alright here's the problem. Also it is a definite integral from A=0 and b = 1
\int (x^7)(x^4-1)^{1/3}

Any help is good.
 
Physics news on Phys.org
Well, what have you tried? There is only one subsitution that would make any sense and that is u= x4- 1. What do you get if you try that substitution?
 
HallsofIvy said:
A pretty obvious substitution to try is u= x4- 1. Then du= 4x3dx so you can separate that x7 into x4 x3, use the "x3 with dx (and never write an integral without the "dx") and then x4= u+ 1.
\int x^7(x^4-1)dx= \int (u+1)u^{\frac{1}{3}}du= \int (u^{\frac{4}{3}}+ u^{\frac{1}{3}}*du

Ok right, exactly. And the new a and b would become a = -1 and b = 0 correct?
 
Thanks, I got it now!1
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top