How to integrate $\sec^{-1}\sqrt\frac{x}{a-x}$

Question

$\int\sec^{-1}\sqrt\frac{x}{a-x}dx$

How to solve the above integration? Please give me some hint about what substitution I should make.

Thanks in advance.

Answer 1

\begin{align} I&=\int\sec^{-1}\sqrt\frac{x}{a-x}\ dx,\quad \color{red}{\sqrt\frac{x}{a-x}=y}\\ &=\int\sec^{-1}(y)\frac{2ay}{(1+y^2)^2}\ dy\quad \color{red}{\text{apply IBP}}\\ &=-\frac{a\sec^{-1}(y)}{1+y^2}+a\int\frac{dy}{y(1+y^2)\sqrt{y^2-1}},\quad \color{red}{y=\sec\theta}\\ &=-\frac{a\sec^{-1}(y)}{1+y^2}+a\int\frac{d\theta}{1+\sec^2\theta},\quad \color{red}{y=\sec\theta}\\ \end{align}

Lets take care of the last integral: \begin{align} K&=\int\frac{d\theta}{1+\sec^2\theta}\\ &=\int\frac{1}{2+\tan^2\theta}\ d\theta, \quad\color{red}{1=\sec^2\theta-\tan^2\theta}\\ &=\int\frac{\sec^2\theta-\tan^2\theta}{2+\tan^2\theta}\ d\theta\\ &=\int\frac{\sec^2\theta}{2+\tan^2\theta}\ d\theta-\int\frac{\tan^2\theta\color{red}{+2-2}}{2+\tan^2\theta}\ d\theta\\ &=\frac1{\sqrt{2}}\tan^{-1}\left(\frac{\tan\theta}{\sqrt{2}}\right)-\int\left(1-\frac{2}{2+\tan^2\theta}\ d\theta\right)\\ &=\frac1{\sqrt{2}}\tan^{-1}\left(\frac{\tan\theta}{\sqrt{2}}\right)-\theta+2K\\ -K&=\frac1{\sqrt{2}}\tan^{-1}\left(\frac{\tan\theta}{\sqrt{2}}\right)-\theta\\ \end{align}

Answer 2

$$\int \:arcsec\left(\frac{\sqrt{x}}{\sqrt{a-x}}\right)dx$$

Put $u=\sqrt{x}$ so $\frac{du}{dx}=\frac{1}{2\sqrt{x}}$ so $dx=2\sqrt{x}du$

$$2\int \:u\:arcsec\left(\frac{u}{\sqrt{a-u^2}}\right)du$$

Integrate by parts $∫fg′= fg−∫f′g$

NOTE-Ignore the factor of $2$ and continue with the integral, we will include in end $$f=arcsec\left(\frac{u}{\sqrt{a-u^2}}\right)$$ and $g$′$=u$

$$f′=\frac{\sqrt{a-u^2}\left(\frac{1}{\sqrt{a-u^2}}+\frac{u^2}{\left(a-u^2\right)^{\frac{3}{2}}}\right)}{u\sqrt{\frac{u^2}{a-u^2}-1}}$$

$$g=\frac{u^2}{2\:}$$

$$=\frac{u^2\:arcsec\left(\frac{u}{\sqrt{a-u^2}}\right)}{2}-\int \:\frac{u\sqrt{a-u^2}\left(\frac{1}{\sqrt{a-u^2}}+\frac{u^2}{\left(a-u^2\right)^{\frac{3}{2}}}\right)}{2\sqrt{\frac{u^2}{a-u^2}-1}}du$$

Now consider the integral in second term only Put $v=u^2$ so $\frac{dv}{du}=2u$ and $du=\frac{1}{2u}dv$

$$\frac{a}{2}\int \:\frac{u}{\left(a-u^2\right)\sqrt{\frac{u^2}{a-u^2}-1}}du$$

$$\frac{a}{2}\int \frac{1}{\left(a-v\right)\sqrt{\frac{v}{a-v}-1}}dv$$

Put $w=\sqrt{\frac{v}{a-v}-1}$ so $\frac{dw}{\:dv}=\frac{\frac{a}{\left(a-v\right)^2}}{2\sqrt{\frac{v}{a-v}-1}}$ and $w^2+1=\frac{v}{a-v}$

$$\frac{a}{2}\int \:\frac{\left(a-v\right)}{a}dw$$

Put $w^2+2=\frac{a}{a-v}$

$$=\frac{a}{2}\int \:\frac{1}{w^2+2}dw$$

This is a standard integral $\int \:\frac{1}{y^2+1}dy=arctan\left(y\right)$

$$\left(\frac{a}{2}\right)\frac{arctan\left(\frac{w}{\sqrt{2}}\right)}{\sqrt{2}}$$

Put back $w=\sqrt{\frac{v}{a-v}-1}$ and $v=u^2$

$$\frac{a\:arctan\left(\frac{\sqrt{\frac{u^2}{a-u^2}-1}}{\sqrt{2}}\right)}{2^{\frac{3}{2}}}$$

So combining this with the previously calculated first term

$$\frac{u^2\:arcsec\left(\frac{u}{\sqrt{a-u^2}}\right)}{2}-\frac{a\:arctan\left(\frac{\sqrt{\frac{u^2}{a-u^2}-1}}{\sqrt{2}}\right)}{2^{\frac{3}{2}}}$$

Put back $u=\sqrt{x}$ and multiply the missing factor of $2$ (See note)

So the final answer is

$$\:x\:arcsec\left(\frac{\sqrt{x}}{\sqrt{a-x}}\right)-\frac{a\:arctan\left(\frac{\sqrt{\frac{x}{a-x}-1}}{\sqrt{2}}\right)}{\sqrt{2}}+ C$$