How to evaluate this indefinite integral by only using trigonometric substitution?

Question

How to evaluate $$\int \sqrt{\frac x{x-1}}dx?$$ I am trying to evaluate this integral by only using trigonometric substitution. I am thinking of substituting $x=\sec^2\theta$. Therefore, the above integral will become $$2\int\sqrt{\frac{\sec^2\theta}{\tan^2\theta}}\sec^2\theta\tan\theta d\theta.$$ Therefore, after this step, we can write the above integral as $$2 \int \sqrt{\operatorname{cosec}^2\theta}\sec^2\theta\tan\theta d\theta.$$ Now here one interesting fact which I noticed is that we can't write $\sqrt{\operatorname{cosec}^2\theta}=|\operatorname{cosec}\theta|$. The only reason for this is that we can't write $|\operatorname{cosec}\theta|=\operatorname{cosec}\theta$ because $|\operatorname{cosec}\theta|=\operatorname{cosec}\theta$ only when $\operatorname{cosec}\theta>0$. But we know that the range of $\operatorname{cosec}\theta$ is $(-\infty,-1]\cup[1,\infty)$. And $\operatorname{cosec}\theta>0$ means $\operatorname{cosec}\theta\in(0,1]\cup[1,\infty)$. Therefore, $|\operatorname{cosec}\theta|\ne\operatorname{cosec}\theta$. Similarly $|\operatorname{cosec}\theta|\neq-\operatorname{cosec}\theta$ and also $|\operatorname{cosec}\theta|\ne0$. Therefore I am facing problem in integrating $$2\int\sqrt{\operatorname{cosec}^2\theta}\sec^2\theta\tan\theta d\theta.$$ Please help me out with this integral.

Answer 1

You may want to render directly

$u=\sqrt{\dfrac{x}{x-1}}$

Then

$u^2=\dfrac{x}{x-1}$

$xu^2-u^2=x$

$x=\dfrac{u^2}{u^2-1}$

Then

$dx=-\dfrac{2u}{(u^2-1)^2}du$

So the full integrand is rendered as

$u dx=-\dfrac{2u^2}{(u^2-1)^2}du$

which should be handled by partial fractions.

Answer 2

Let $u=\sqrt{\frac{x}{x-1}} $, then $x=\frac{u^2}{u^2-1}=1+\frac{1}{u^2-1} $ and $d x=d\left(\frac{1}{u^2-1}\right)$.

By integration by parts, we have $$ \begin{aligned} I & =\int u d\left(\frac{1}{u^2-1}\right) \\ & =\frac{u}{u^2-1}-\int \frac{1}{u^2-1} d u \\ & =\frac{u}{u^2-1}-\frac12\ln \left|\frac{u-1}{u+1}\right|+C\\&=(x-1) \sqrt{\frac{x}{x-1}} +\frac12\ln \left|\frac{\sqrt{\frac{x}{x-1}}+1}{\sqrt{\frac{x}{x-1}}-1}\right|+C \end{aligned} $$ \begin{aligned} & \end{aligned}

Answer 3

You did well by setting $x=\sec^2\theta$ and already got

$\qquad I=\int \sqrt{\frac x{x-1}}dx=2\int \sqrt{\operatorname{cosec}^2\theta}\sec^2\theta\tan\theta d\theta$.

Note that $x$ must be greater than $1$, otherwise the integrand will NOT be a real number. Now simply continue to go ahead without your worry to get

$\qquad I=2\int cosec\ \theta\sec^2\theta\tan\theta\ d\theta=2\int sec^3\theta\ d\theta$.

This will lead to

$\qquad I=\frac{sin\theta}{cos^2\theta}+ln(\frac{1+sin\theta}{cos\theta})+c$,

where $c$ is any constant. Since $\theta=arcsec(\sqrt{x})$, so we have

$\qquad sin\theta=\sqrt{\frac{x-1}{x}}$, $cos\theta=\sqrt{\frac{1}{x}}$.

Substituting these into the expression of $I$ finally yields

$\qquad I=\sqrt{x(x-1)}+ln(\sqrt{x}+\sqrt{x-1})+c$.

The correctness of this final expression can be verified by deriving the derivative $\frac{dI}{dx}$ to see that indeed $\frac{dI}{dx}$ equals the original integrand.