Indefinite Integration - with trignometric substitution

Question

$$\int \sqrt{\frac{\sqrt{x}-1}{ \sqrt{x}+1}} \,\,dx$$

The tip I got was to use $x= \sec^2(a)$. I don't no how to continue. I think there is trignometric simplification.

Answer 1

\begin{align*} x &= \sec^2 t \\ dx &= 2\sec^2 t \tan t \, dt \\ \sqrt{\frac{\sqrt{x}-1}{\sqrt{x}+1}} &= \sqrt{\frac{(\sqrt{x}-1)(\sqrt{x}-1)}{(\sqrt{x}+1)(\sqrt{x}-1)}} \\ &= \frac{\sqrt{x}-1}{\sqrt{x-1}} \\ &= \frac{\sec t-1}{\tan t} \\ I &= \int \sqrt{\frac{\sqrt{x}-1}{\sqrt{x}+1}} \, dx \\ &= \int 2\sec^2 t(\sec t-1) \, dt \\ &= 2\int \sec^3 t\, dt-2\int \sec^2 t\, dt \\ &= 2\int \sec^3 t\, dt-2\tan t \\ \int \sec^3 t\, dt &= \int \sec t\, d(\tan t) \\ &= \sec t \tan t-\int \tan t\, d(\sec t) \\ &= \sec t \tan t-\int \tan^2 t \sec t\, dt \\ &= \sec t \tan t-\int (\sec^2 t-1) \sec t\, dt \\ &= \sec t \tan t-\int \sec^3 t\, dt+\int \sec t\, dt \\ 2\int \sec^3 t\, dt &= \sec t \tan t+\int \sec t\, dt \\ \int \sec t\, dt &= \int \frac{\sec t \tan t+\sec^2 t}{\sec t+\tan t}\, dt \\ &= \int \frac{d(\sec t+\tan t)}{\sec t+\tan t} \\ &= \ln |\sec t+\tan t|+C \\ I &= (\sec t-2)\tan t+\ln |\sec t+\tan t|+C \\ &= (\sqrt{x}-2)\sqrt{x-1}+\ln (\sqrt{x}+\sqrt{x-1})+C \end{align*}

Answer 2

HINT: set $$t=\sqrt{\frac{\sqrt{x}-1}{\sqrt{x}+1}}$$ then we have $$x=\left(\frac{t^2+1}{t^2-1}\right)^2$$ and $$dx=-\frac{8(t^2+1)t}{(t-1)^3(t+1)^3}dt$$

Answer 3

hint

put $x=\cos^2(2t) $ and use

$$1-\cos(2t)=2\sin^2(t) $$

$$1+\cos(2t)=2\cos^2(t) .$$

Answer 4

$$\int \sqrt{\frac{\sqrt{x}-1}{ \sqrt{x}+1}} \,dx=\int \frac{\sqrt{x}-1}{\sqrt{ x-1}}\,dx.$$

The term $-1$ at the numerator integrates immediately.

Then with $x=\cosh^2t$,

$$\int \frac{\sqrt{x}}{\sqrt{ x-1}}\,dx=2\int \frac{\cosh t}{\sinh t}\cosh t\sinh t\,dt=\int(\cosh 2t+1)\,dt.$$