Gdtyjr

I would say I'm rather good at doing trig substitution when there is a square root, but when there isn't one, I'm lost.

I'm currently trying to solve the following question:

$Ar int_a^infty fracdx(r^2+x^2)^(3/2)$

Anyway, so far, I have that:

$x = rtan theta$

$dx = rsec^2 theta$

$sqrt (r^2+x^2) = rsectheta$

Please click here to see the triangle I based the above values on.

.

Given that $(r^2+x^2)^(3/2)$ can be rewritten as $ (sqrtr^2+x^2)^3$, I begin to solve.
Please pretend I have $lim limits_b to infty$ in front of every line please.

= $Ar int_a^b fracrsec^2theta(rsectheta)^3dtheta$

= $Ar int_a^b fracrsec^2thetar^3sec^6thetadtheta$

= $fracAr int_a^b frac1sec^4thetadtheta$

= $fracAr int_a^b cos^4theta dtheta$

= $fracAr int_a^b (cos^2theta)^2 dtheta$

= $fracAr int_a^b [ frac121+cos(2theta)) ]^2dtheta$

= $fracA4r int_a^b 1 + 2cos(2theta) + cos^2(2theta) dtheta$

= $fracA4r int_a^b 1 + 2cos(2theta) dtheta quad+quad fracA4r int_a^b cos^2(2theta) dtheta$

And from there it gets really messed up and I end up with a weird semi-final answer of $fracA4r[2theta+sin(2theta)] + fracA32r [4theta+sin(4theta)]$ which is wrong after I make substitutions.

.

I've done this a couple different ways (I keep messing up), but this is the most correct that I've been able to come up with and it's still wrong.

It would be soooo awesome if someone could tell me if I'm even on the right track. Like I said, if it was a simple square root instead of $quad ^3/2quad$, life would be awesome.

I already know that the final answer is $fracAr(1-fracasqrtr^2+a^2)$, but I really want to understand this. An explanation would be most welcome!

Thank you in advance for all your help!

You are doing $(rsectheta)^3=r^6sec^6theta$. Oops! ;-)

There's a slicker way to do it.

Get rid of the $r$ with $x=ru$ to begin with, so your integral becomes
$$
fracArint_a/r^inftyfrac1(1+u^2)^3/2,du
$$
Now let's concentrate on the antiderivative
$$
intfrac1(1+u^2)^3/2,du=
intfrac1+u^2-u^2(1+u^2)^3/2,du=
intfrac1(1+u^2)^1/2,du-intfracu^2(1+u^2)^3/2,du
$$
Do the second term by parts
$$
int ufracu(1+u^2)^3/2,du=
-fracu(1+u^2)^1/2+intfrac1(1+u^2)^1/2,du
$$
See what happens?
$$
intfrac1(1+u^2)^3/2,du=fracu(1+u^2)^1/2+c
$$
which we can verify by direct differentiation.

Now
$$
left[fracu(1+u^2)^1/2right]_a/r^infty=1-fraca/r(1+(a/r)^2)^1/2
=1-fraca(r^2+a^2)^1/2
$$
and your integral is indeed
$$
fracArleft(1-fracasqrtr^2+a^2right)
$$

Firstly you made an error in the first line of working
$$(rsec(theta))^3=r^3sec^3(theta)$$
Secondly, you need to change the range of integration after performing a substitution. If $theta=arctan(fracxr)$ then the limits should change as $x=a implies theta=arctan(fracar)$ also $x=infty implies theta=fracpi2$.