Question

anti-derivative of x^3sqrt(x^2-1)
using a u substitution

Answer 1

first thing you should try is to sub the stuff inside radical

Answer 2

sub \(\large u = x^2-1\)

Answer 3

did that

Answer 4

take du=2x dx, dx= 1/2x du

Answer 5

yes, we can do better :

\(\large u = x^2 - 1 \implies x^2 = u-1\)
\(\large du = 2x dx\)
\(\large xdx = \frac{1}{2}du\)

Answer 6

\(\large \int x^3\sqrt{x^2-1}~ dx\)

\(\large \int x^2. x \sqrt{x^2-1}~ dx\)

\(\large \int x^2 \sqrt{x^2-1} ~xdx\)

Answer 7

substitute now ^

Answer 8

the integral becomes :
\(\large \int (u+1)\sqrt{u} ~ \dfrac{du}{2}\)

Answer 9

should be trivial to evaluate ?

Answer 10

should that not be du/2x

Answer 11

that x got stuck with the "dx" on the left hand side itself

Answer 12

ok so, now what? this is exactly were i got stuck on my own lol. do i sub back for u to get an equation in the form of x^2sqrt(x^2-1)du/2?

Answer 13

\(\large \int (u+1)\sqrt{u} ~ \dfrac{du}{2}\)

Answer 14

you're fine till that step ?

Answer 15

yes

Answer 16

multiply

Answer 17

\(\large \dfrac{1}{2}\int (u\sqrt{u}+\sqrt{u} )~ du\)

Answer 18

\(\large \dfrac{1}{2}\int (u^{\frac{3}{2}}+u^{\frac{1}{2}})~ du\)

Answer 19

use below formula to evaluate :
\[\large \int x^n dx = \dfrac{x^{n+1}}{n+1}+C\]

Answer 20

is that a calc 2 formula? this is calc 1 and i dont believe we were given that to evaluate our types of intergrals

Answer 21

thats calc1 formula

Answer 22

ok ok thats familiar

Answer 23

good :)

Answer 24

by doing that i get 1/2[(x^2-1)^5/2)/5/2+(x^2-1)^3/2)/3/2))

Answer 25

looks good ! don't forget the constant C

Answer 26

the correct answer according to wolfram was supposed to be
\[[(x ^{2}-1)^3/2 *(3x^2 +2)]/15 \]

Answer 27

both are same
http://www.wolframalpha.com/input/?i=simplify++1%2F2%28%28x%5E2-1%29%5E%285%2F2%29%2F%285%2F2%29%2B%28x%5E2-1%29%5E%283%2F2%29%2F%283%2F2%29%29

Answer 28

thanks so much. ill be posting another similar substitution in a few minutes once i jot this down in my note book

Answer 29

np :)