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:
$$Ar \int_a^\infty \frac{dx}{(r^2+x^2)^{(3/2)}}$$
Anyway, so far, I have that:
$$x = r\tan \theta$$
$$dx = r\sec^2 \theta$$
$$\sqrt {(r^2+x^2)} = r\sec\theta$$
The triangle I based the above values on:
Given that $(r^2+x^2)^{(3/2)}$ can be rewritten as $ (\sqrt{r^2+x^2})^3$, I begin to solve.
Please pretend I have $\lim \limits_{b \to \infty}$ in front of every line please.
\begin{align}
&= Ar \int_a^b \frac{r\sec^2\theta}{(r\sec\theta)^3}d\theta \\
&= Ar \int_a^b \frac{r\sec^2\theta}{r^3\sec^6\theta}d\theta \\
&= \frac{A}{r} \int_a^b \frac{1}{\sec^4\theta}d\theta \\
&= \frac{A}{r} \int_a^b \cos^4\theta d\theta \\
&= \frac{A}{r} \int_a^b (\cos^2\theta)^2 d\theta \\
&= \frac{A}{r} \int_a^b \left[\ \frac12 1+\cos(2\theta))\ \right]^2d\theta \\
&= \frac{A}{4r} \int_a^b 1 + 2\cos(2\theta) + \cos^2(2\theta)\ d\theta \\
&= \frac{A}{4r} \int_a^b 1 + 2\cos(2\theta)\ d\theta \quad+\quad \frac{A}{4r} \int_a^b \cos^2(2\theta)\ d\theta
\end{align}
And from there it gets really messed up and I end up with a weird semi-final answer of $$\frac{A}{4r}[2\theta+\sin(2\theta)] + \frac{A}{32r} [4\theta+\sin(4\theta)]$$ which is wrong after I make substitutions.
I already know that the final answer is $\dfrac{A}{r}\left(1-\dfrac{a}{\sqrt{r^2+a^2}}\right)$, but I really want to understand this.
Best Answer
You are doing $(r\sec\theta)^3=r^6\sec^6\theta$. Oops!
;-)There's a slicker way to do it.
Get rid of the $r$ with $x=ru$ to begin with, so your integral becomes $$ \frac{A}{r}\int_{a/r}^{\infty}\frac{1}{(1+u^2)^{3/2}}\,du $$ Now let's concentrate on the antiderivative $$ \int\frac{1}{(1+u^2)^{3/2}}\,du= \int\frac{1+u^2-u^2}{(1+u^2)^{3/2}}\,du= \int\frac{1}{(1+u^2)^{1/2}}\,du-\int\frac{u^2}{(1+u^2)^{3/2}}\,du $$ Do the second term by parts $$ \int u\frac{u}{(1+u^2)^{3/2}}\,du= -\frac{u}{(1+u^2)^{1/2}}+\int\frac{1}{(1+u^2)^{1/2}}\,du $$ See what happens? $$ \int\frac{1}{(1+u^2)^{3/2}}\,du=\frac{u}{(1+u^2)^{1/2}}+c $$ which we can verify by direct differentiation.
Now $$ \left[\frac{u}{(1+u^2)^{1/2}}\right]_{a/r}^{\infty}=1-\frac{a/r}{(1+(a/r)^2)^{1/2}} =1-\frac{a}{(r^2+a^2)^{1/2}} $$ and your integral is indeed $$ \frac{A}{r}\left(1-\frac{a}{\sqrt{r^2+a^2}}\right) $$