How to integrate $\int \sqrt{1-\dfrac{1}{25x^2}}\ dx$?

Question

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

Answer 1

Your answer is correct.

You can use another substitution $\dfrac{1}{5x}=\sin\theta\implies dx=\dfrac{-\cos\theta\ d\theta}{5\sin^2\theta}$

$$\int \sqrt{1-\dfrac{1}{25x^2}}\ dx=\int \cos\theta \left( \dfrac{-\cos\theta\ d\theta}{5\sin^2\theta}\right)$$ $$=\frac15\int (-\cot^2\theta) \ d\theta$$ $$=\frac15\int (1-\csc^2\theta) \ d\theta$$ $$=\frac15(\theta+\cot\theta)+C$$ $$=\frac15\sin^{-1}\left(\frac{1}{5x}\right)+\frac15\sqrt{25x^2-1}+C$$

Answer 2

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 $t=\sqrt{25x^2-1}$.

Answer 3

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) $,

$$=F(x)=-\frac 15\int \frac{u^2}{1+u^2}du$$ $$=-\frac 15(u-\arctan(u))+C$$

$$=-\frac 15\Bigl(\sqrt{25x^2-1}-\arctan(\sqrt{25x^2-1})\Bigr)+C$$

Answer 4

It is more convenient to just integrate by parts \begin{align} \int \sqrt{1-\frac{1}{25x^2}}\ dx &= 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}

Answer 5

You are right. Differentiate the solution to get original function $$\dfrac{d}{dx}\left( \dfrac15\sqrt{25x^2-1}-\frac15\sec^{-1}(5x)\right)=\sqrt{1-\dfrac{1}{25x^2}}$$

$$\int \sqrt{1-\dfrac{1}{25x^2}}\ dx=\dfrac15\sqrt{25x^2-1}-\frac15\sec^{-1}(5x)+c$$