Trigonometric Substitution (integration) $$\int\frac{dx}{(x^{2}-36)^{3/2}}$$
My attempt:
the factor in the denominator implies
$$x^{2}-36=x^{2}-6^{2}$$
substituting $x=6\sec\theta$, noting that $dx=6\tan\theta \sec\theta$ 
$$x^{2}-6^{2}=6^{2}\sec^{2}\theta-6^{2}=6^{2}\tan^{2}\theta$$
$$\int\frac{dx}{(x^{2}-36)^{3/2}}=\int\frac{6\tan\theta \sec\theta}{36\tan^{2}\theta}=\frac{1}{6}\int\frac{\sec\theta}{\tan\theta}$$
using trig identities:
$$\frac{1}{6}\int\frac{\sec\theta}{\tan\theta}=\frac{1}{6}\int \sin^{-1}\theta$$
now using integration by parts:
$$\frac{1}{6}\int \sin^{-1}\theta$$
$$u=\sin^{-1}\theta, du=\frac{1}{\sqrt{1-\theta^{2}}}, dv=1, v=\theta$$
using $uv-\int{vdu}$
$$\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\int{\frac{\theta}{\sqrt{1-\theta^{2}}}}d\theta\bigg)$$
now using simple substitution:$$z=1-\theta^{2}, dz=-2\theta d\theta, -\frac{1}{2}du=\theta d\theta$$
it is apparent that
$$\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\bigg(-\frac{1}{2}\int{\frac{dz}{\sqrt{z}}}\bigg)\bigg)$$
$$=\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\bigg(-\frac{1}{2}(2\sqrt{z})\bigg)\bigg)=\frac{1}{6}\bigg(\theta \sin^{-1}\theta+\sqrt{1+\theta^{2}}\bigg)$$
$$=\frac{1}{6}\theta \sin^{-1}\theta+\frac{1}{6}\sqrt{1+\theta^{2}}+C$$
I have the following questions:
1.This integral seems tricky and drawn out to me, is there another method that reduces the steps/ methods of integration? I had to use trig substitution, integration by parts, and substitution in order to solve the integral, what can I do to find easier ways to complete integrals of this type?
2.Is this solution even correct? wolfram alpha says the solution to this integral is $-\frac{x}{36\sqrt{x^{2}-36}}+C$ how can i determine equivalence?
 A: Be careful: with the substitution you have
$$
(x^2-36)^{3/2}=(36\tan^2\theta)^{3/2}=216\tan^3\theta
$$
(at least in an interval where $\tan\theta$ is positive) so your integral becomes
$$
\int\frac{6\tan\theta\sec\theta}{216\tan^3\theta}\,d\theta=
\frac{1}{36}\int\frac{\cos\theta}{\sin^2\theta}\,d\theta=
-\frac{1}{36}\frac{1}{\sin\theta}+C
$$
If instead you set $x=6\cosh u$, you get $dx=6\sinh u$ and the identity $\cosh^2u-1=\sinh^2u$ brings the integral in the form
$$
\frac{1}{36}\int\frac{1}{\sinh^2u}\,du=
-\frac{1}{36}\frac{\cosh u}{\sinh u}+C=
-\frac{1}{36}\frac{x}{\sqrt{x^2-36}}
$$
A: Without substitution:
$$\int\frac{dx}{(x^{2}-36)^{3/2}}=\frac1{36}\int\frac{(x^2-(x^2-36))\,dx}{(x^{2}-36)^{3/2}}=\frac1{36}\int\frac{x^2\,dx}{(x^{2}-36)^{3/2}}-\frac1{36}\int\frac{dx}{(x^{2}-36)^{1/2}}.$$
Then by parts on the first term,
$$\int\frac{x\cdot x\,dx}{(x^{2}-36)^{3/2}}=-\frac x{(x^{2}-36)^{1/2}}+\int\frac{dx}{(x^{2}-36)^{1/2}}.$$
After scaling the variable with a factor $6$, you recognize a known derivative,
$$\int\frac{dx}{\sqrt{x^2-1}}=\text{arcosh}(x).$$
A: Your solution is incorrect as $$\frac{\sec\theta}{\tan\theta}$$ is not the same as $$\ sin^{-1}\theta$$ It is however, equal to $\csc\theta$, a standard integral that you had got after a few steps
A: attempt 2:
$$\int\frac{dx}{(x^{2}-36)^{3/2}}$$
the factor in the denominator implies
$$(x^{2}-36)^{3/2}=({x^{2}-6^{2}})^{3/2}$$
substituting $x=6sec\theta$, noting that $dx=6tan\theta sec\theta d\theta$ 
$$(x^{2}-6^{2})^{3/2}=(6^{2}sec^{2}\theta-6^{2})^{3/2}d\theta=(36tan^{2}\theta)^{3/2}=216tan^3\theta$$
$$\int\frac{dx}{(x^{2}-36)^{3/2}}=\int\frac{6tan\theta sec\theta}{216tan^{3}\theta}d\theta=\frac{1}{36}\int\frac{sec\theta}{tan^{2}\theta}d\theta$$
using trig identities:
$$\frac{1}{36}\int\frac{sec\theta}{tan^{2}\theta}d\theta=\frac{1}{36}\int{cot(\theta)csc(\theta) d\theta}$$
the integral cot(x)csc(x) is know
$$\int cot(x)sec(x)dx=-csc(x)+C$$
so 
$$\frac{1}{36}\int cot(\theta)csc(\theta)d\theta=\frac{1}{36}\bigg(-csc(\theta)\bigg)+C=-\frac{1}{36}csc(\theta)+C$$
Question 2 still applies:
2.Is this solution even correct? wolfram alpha says the solution to this integral is $-\frac{x}{36\sqrt{x^{2}-36}}+C$ how can i determine equivalence?
