What technique should I use to solve this one?

int(16+x^2)^(-3/2)

I tried a few techniques, including partial fractions, but I couldn't work it out. Can anybody help me?

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- Mar 18th 2007, 10:06 AMriverjibIntegration technique help, please!
What technique should I use to solve this one?

int(16+x^2)^(-3/2)

I tried a few techniques, including partial fractions, but I couldn't work it out. Can anybody help me? - Mar 18th 2007, 10:25 AMtopsquark
- Mar 18th 2007, 10:54 AMJhevon
I agree with topsquark, trig substitution seems like the way to go.

you have int(16+x^2)^(-3/2) = int((4)^2+x^2)^(-3/2) = int[1/(sqrt((4)^2 + x^2)^3].

trig substitution says, if your integral contains sqrt(a^2 + x^2), where a is a constant, you use the sustitution, x = atan@ and the identity 1 + tan^2@ = sex^2@ - Mar 18th 2007, 11:11 AMtopsquark
- Mar 18th 2007, 11:17 AMJhevon
- Mar 18th 2007, 01:22 PMriverjib
Thank you both! That makes perfect sense.

- Mar 18th 2007, 06:09 PMriverjibNeed a bit more help...sorry
I started by using the substitution x=4tan@, but how do i handle the fact that a^2+x^2 is cubed?? none of the trig substitution problems we've done have added that kind of complication.

- Mar 18th 2007, 10:54 PMtopsquark
Let x = 4*tan(t) ==> dx = 4*sec^2(t) dt

Then

Int[(16+x^2)^(-3/2)dx] =

Int[(16+16{tan^2(t)})^(-3/2) * 4sec^2(t) dt]

= 4*(16)^(-3/2)*Int[(1 + tan^2(t))^(-3/2) sec^2(t) dt]

= 4*(1/64)*Int[(sec^2(t))^(-3/2) sec^2(t) dt]

= (4/64)*Int[sec^2(t)/sec^3(t) dt]

= (1/16)*Int[1/sec(t) dt]

= (1/16)*Int[cos(t) dt]

= (1/16)*sin(t)

Now, t = tan^{-1}(x/4), thus

Int[(16+x^2)^(-3/2)dx] = (1/16)*sin[tan^{-1}(x/4)]

(plus some constant of integration.)

Now, imagine a right triangle with legs x and 4, then the hypotenuse will be sqrt(x^2 + 16). We have an angle t (which is tan^{-1}(x/4)) across from the side of length x. Then the sine of this angle will be: x/sqrt(x^2 + 16). So

sin[tan^{-1}(x/4)] = x/sqrt(x^2 + 16)

So finally:

Int[(16+x^2)^(-3/2)dx] = (1/16)*x/sqrt(x^2 + 16)

(Note: I finally found the error in my original post. This is now correct.)

-Dan - Mar 19th 2007, 12:52 PMriverjibThank you so much!
Thanks Dan! I read over your solution, gave it a few hours, and then worked it out on my own. Now I really understand every trig substitution problem in the book. I'm ready for my exam :-)

I really appreciate all your help. - Mar 20th 2007, 05:02 AMtopsquark