# Thread: ASAP Calculus Integration help (by parts and trig)

1. ## 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?

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

integration by parts ...

$\displaystyle u = x^2$

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

$\displaystyle du = 2x \, dx$

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

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

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

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

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

3. Hello, valkyrie!

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

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

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

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*
.

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

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 $\displaystyle u = \sec{t}$

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

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

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

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

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

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

5. 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?

6. 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 $\displaystyle \frac 12 \int (u^{1/2} - u^{-1/2})~du$

7. thanks a lot! this was really helpful!