How to integrate $\int \sqrt{1-\dfrac{1}{25x^2}}\ dx$? How to integrate the following:
$$\int \sqrt{1-\dfrac{1}{25x^2}}\ dx$$
What I did is:
$$\int \sqrt{1-\dfrac{1}{25x^2}}\ dx=\int \dfrac{\sqrt{25x^2-1}}{5x}\ dx$$
I substituted $5x=\sec\theta$, $dx=\dfrac{1}{5}\sec\theta\tan\theta\ d\theta $
$$=\int \dfrac{\sqrt{\sec^2\theta-1}}{\sec\theta}\ \dfrac{1}{5}\sec\theta\tan\theta\ d\theta$$
$$=\frac15\int \tan^2\theta\ d\theta$$
used $\tan^2\theta=\sec^2\theta-1$
$$=\frac15\int( \sec^2\theta-1)\ d\theta$$
$$=\dfrac15\tan\theta-\frac15\theta+c$$
back to $x$
$$=\dfrac15\sqrt{25x^2-1}-\frac15\sec^{-1}(5x)+c$$
I am not sure whether my answer is correct.
My question: Can I integrate this with other substitutions? If yes, please help me. Thank you
 A: Your answer is correct.
You can use another substitution
$\dfrac{1}{5x}=\sin\theta\implies dx=\dfrac{-\cos\theta\ d\theta}{5\sin^2\theta}$
$$\int \sqrt{1-\dfrac{1}{25x^2}}\ dx=\int \cos\theta \left( \dfrac{-\cos\theta\ d\theta}{5\sin^2\theta}\right)$$
$$=\frac15\int (-\cot^2\theta) \ d\theta$$
$$=\frac15\int (1-\csc^2\theta) \ d\theta$$
$$=\frac15(\theta+\cot\theta)+C$$
$$=\frac15\sin^{-1}\left(\frac{1}{5x}\right)+\frac15\sqrt{25x^2-1}+C$$
A: you can also do it with $t^2=25x^2-1$. To get
\begin{align*}
\int \frac{\sqrt{25x^2-1}}{5x} \, dx&=\frac{1}{5}\int \frac{t^2}{t^2+1} dt\\
&=\frac{1}{5}\left[\int \frac{t^2+1-1}{t^2+1} dt\right]\\
&=\frac{1}{5}\left[t-\int \frac{1}{t^2+1} dt\right]\\
&=\frac{t}{5}-\frac{\arctan t}{5}+c,
\end{align*}
where $t=\sqrt{25x^2-1}$.
A: The domain of the integrand is
$$(-\infty,-\frac 15]\cup [\frac 15,+\infty)$$
Assume we want the antiderivative at $(-\infty,-\frac 15]$,
$$F(x)=\int \sqrt{\frac{25x^2-1}{25x^2}}dx=$$
$$-\frac 15\int \frac{\sqrt{25x^2-1}}{x}dx$$
Now you can put
$$5x=-\cosh(t)$$
with
$$\sqrt{25x^2-1}=\sqrt{\cosh^2(t)-1}=\sinh(t)$$
then
$$F(x)=-\int \frac{\sinh(t)}{\cosh(t)}\frac{\sinh(t)dt}{5}$$
which becomes, with $ u=\sinh(t) $,
$$=F(x)=-\frac 15\int \frac{u^2}{1+u^2}du$$
$$=-\frac 15(u-\arctan(u))+C$$
$$=-\frac 15\Bigl(\sqrt{25x^2-1}-\arctan(\sqrt{25x^2-1})\Bigr)+C$$
A: Integrate over all domain $x$ as follows\begin{align}
\int \sqrt{1-\frac{1}{25x^2}}\ dx
&\overset{ibp}= x \sqrt{1-\frac{1}{25x^2}} + \frac1{5} \int \frac1{\sqrt{1-\frac{1}{25x^2}}}d(\frac1{5x})\\
&= x \sqrt{1-\frac{1}{25x^2}} +\frac15 \sin^{-1}\frac1{5x}+C
\end{align}
A: You are right. Differentiate the solution to get original function
$$\dfrac{d}{dx}\left( \dfrac15\sqrt{25x^2-1}-\frac15\sec^{-1}(5x)\right)=\sqrt{1-\dfrac{1}{25x^2}}$$
$$\int \sqrt{1-\dfrac{1}{25x^2}}\ dx=\dfrac15\sqrt{25x^2-1}-\frac15\sec^{-1}(5x)+c$$
