<s>[INST] 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
 [/INST] 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 $t=\sqrt{25x^2-1}$.
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