Ttyjfyk

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:

$x = rtan theta$

$dx = rsec^2 theta$

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

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 already know that the final answer is $fracAr(1-fracasqrtr^2+a^2)$, but I really want to understand this.

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 $x=infty implies theta=fracpi2$.