Evaluate $\int \frac{a}{x(x^2-a^2)^{1/2}} dx$ I was trying to integrate $\int \frac{a}{x(x^2-a^2)^{1/2}} dx$ and by applying on what i saw on the formula of inverse trigonometric functions, there is formula like $\frac{1}{a}Arcsec\frac{u}{a}$ = $\int \frac{du}{u\sqrt(u^2-a^) }$ so my answer is $\frac{1}{a}Arcsec\frac{x}{a}$ but my friend said it should be something like $arcsin?$
 A: To obtain the solution that the OP quotes
$$
\int \frac{a}{x\sqrt{x^2-a^2}}dx
$$
we then have
$$
\int \frac{a}{x^2\sqrt{1-\left(\frac{a}{x}\right)^2}}dx
$$
Here I used the fact
$$
\sqrt{x^2-a^2} = \sqrt{x^2\left(1-\frac{a^2}{x^2}\right)} = \sqrt{x^2}\sqrt{\left(1-\frac{a^2}{x^2}\right)} = x\sqrt{1-\left(\frac{a}{x}\right)^2}
$$
let $u = \frac{a}{x}$
we have
$$
-\int \frac{du}{\sqrt{1-u^2}} = -\arcsin(u) + C
$$
A: $$I=\int\dfrac1{x\sqrt{x^2-a^2}}dx=\int\dfrac x{x^2\sqrt{x^2-a^2}}dx$$
Let $\sqrt{x^2-a^2}=u\implies\dfrac x{\sqrt{x^2-a^2}}dx=du$ and $x^2=u^2+a^2$
$$I=\int\dfrac{du}{u^2+a^2}=\dfrac1a\arctan\dfrac ua+K$$
Now we can prove $$\arctan y=\arcsin\dfrac y{\sqrt{1+y^2}}\text{ OR } \arcsin z=\arctan\dfrac z{\sqrt{1-z^2}}$$
and use $\arcsin a+\arccos a=\dfrac\pi2$ for $|a|\le1$
and $\arccos p=\text{arcsec}\dfrac1p$ for $|p|\le1$
A: Notice, we have $$\int\frac{adx}{x\sqrt{x^2-a^2}}$$ Let $x=a\sec \theta \implies a\sec\theta\tan \theta d\theta=dt$
$$\int\frac{a^2\sec\theta\tan \theta d\theta}{a\sec\theta\sqrt{a^2 \sec^2\theta-a^2}}$$
$$=\int\frac{a^2\sec\theta\tan \theta d\theta}{a^2\sec\theta\tan \theta}$$ $$=\int d\theta=\theta+c_1$$ Now, setting the value of $\theta$ $$=\sec^{-1}\left(\frac{x}{a}\right)+c_1$$ Since, $\sec^{-1}\left(\frac{x}{a}\right)=\cos^{-1}\left(\frac{a}{x}\right)$ $$=\cos^{-1}\left(\frac{a}{x}\right)+c_1$$ $$=\frac{\pi}{2}-\sin^{-1}\left(\frac{a}{x}\right)+c_1$$ $$=-\sin^{-1}\left(\frac{a}{x}\right)+c_2$$ 
Hence, we have $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\int\frac{a}{x\sqrt{x^2-a^2}}dx=\sec^{-1}\left(\frac{x}{a}\right)+c_1=-\sin^{-1}\left(\frac{a}{x}\right)+c_2}}$$
