# ASAP Calculus Integration help (by parts and trig)

• Feb 11th 2009, 01:28 PM
valkyrie
ASAP Calculus Integration help (by parts and trig)
Hi guys, I'm new on the forum and need a little assistance... we were given this to integrate:

/int x^3/sqrt (x^2+1)dx

And were instructed to integrate by trig integration and integration by parts and then show that they are the same. I got the answer with trig integration:

1/3 sqrt (x^2+1)^3 - 1/sqrt (x^2+1) + c

but I can't figure out how to do it by parts. What should I assign as u and dv and why?
• Feb 11th 2009, 02:01 PM
skeeter
Quote:

/int x^3/sqrt (x^2+1)dx
$\displaystyle \int \frac{x^3}{\sqrt{x^2+1}} \, dx$

integration by parts ...

$\displaystyle u = x^2$

$\displaystyle dv = \frac{x}{\sqrt{x^2+1}} \, dx$

$\displaystyle du = 2x \, dx$

$\displaystyle v = \sqrt{x^2+1}$

$\displaystyle x^2 \sqrt{x^2+1} - \int 2x\sqrt{x^2+1} \, dx$

$\displaystyle x^2\sqrt{x^2+1} - \frac{2}{3}\sqrt{(x^2+1)^3} + C$

$\displaystyle \sqrt{x^2+1}\left[x^2 - \frac{2}{3}(x^2+1)\right] + C$

$\displaystyle \sqrt{x^2+1} \left(\frac{x^2-2}{3}\right) + C$
• Feb 11th 2009, 02:03 PM
Soroban
Hello, valkyrie!

It can be done with just by-parts . . .

Quote:

$\displaystyle \int \frac{x^3}{\sqrt{x^2+1}}\,dx$

. . $\displaystyle \begin{array}{ccccccc}u &=& x^2 & & dv &=&(x^2+1)^{-\frac{1}{2}}(x\,dx) \\ du &=& 2x\,dx & & v &=& (x^2+1)^{\frac{1}{2}} \end{array}$

And we have: .$\displaystyle x^2\sqrt{x^2+1} - \int2x(x^2+1)^{\frac{1}{2}}\,dx$

Got it?

Edit: Skeeter beat me to it . . . *sigh*
.
• Feb 11th 2009, 03:18 PM
skeeter
$\displaystyle \int \frac{x^3}{\sqrt{x^2+1}} \, dx$

trig integration ...

$\displaystyle x = \tan{t}$

$\displaystyle dx = \sec^2{t} \, dt$

$\displaystyle \int \frac{\tan^3{t}}{\sqrt{\tan^2{t}+1}} \cdot \sec^2{t} \, dt$

$\displaystyle \int \tan^3{t}\cdot \sec{t} \, dt$

$\displaystyle \int \tan^2{t} \cdot \sec{t}\tan{t} \, dt$

$\displaystyle \int (\sec^2{t} - 1)\sec{t}\tan{t} \, dt$

let $\displaystyle u = \sec{t}$

$\displaystyle \int (u^2 - 1) \, du$

$\displaystyle \frac{u^3}{3} - u + C$

$\displaystyle \frac{\sec^3{t}}{3} - \sec{t} + C$

$\displaystyle \sec{t}\left(\frac{\sec^2{t} - 1}{3} - \frac{2}{3}\right) + C$

$\displaystyle \sqrt{\tan^2{t}+1} \left(\frac{\tan^2{t} - 2}{3}\right) + C$

$\displaystyle \sqrt{x^2 + 1} \left(\frac{x^2-2}{3}\right) + C$
• Feb 11th 2009, 04:15 PM
valkyrie
Quote:

Originally Posted by skeeter
$\displaystyle \int \frac{x^3}{\sqrt{x^2+1}} \, dx$

integration by parts ...

$\displaystyle u = x^2$

$\displaystyle dv = \frac{x}{\sqrt{x^2+1}} \, dx$

$\displaystyle du = 2x \, dx$

$\displaystyle v = \sqrt{x^2+1}$

$\displaystyle x^2 \sqrt{x^2+1} - \int 2x\sqrt{x^2+1} \, dx$

$\displaystyle x^2\sqrt{x^2+1} - \frac{2}{3}\sqrt{(x^2+1)^3} + C$

$\displaystyle \sqrt{x^2+1}\left[x^2 - \frac{2}{3}(x^2+1)\right] + C$

$\displaystyle \sqrt{x^2+1} \left(\frac{x^2-2}{3}\right) + C$

im confused, where does the 2x go from insude the integral?
• Feb 11th 2009, 04:21 PM
Jhevon
Quote:

Originally Posted by valkyrie
im confused, where does the 2x go from insude the integral?

do the integral, you will see where the 2x went (use integration by substitution).

for your general edification, note that the integral can be done using regular substitution, $\displaystyle u = x^2 + 1$ transforms it into $\displaystyle \frac 12 \int (u^{1/2} - u^{-1/2})~du$
• Feb 11th 2009, 07:13 PM
valkyrie
thanks a lot! this was really helpful!