How can we find the integral $\int_{a}^{b} \sqrt{\left(\frac{1}{x}-1\right)}dx$, where $a, b$ both are positive. I tried to use method of substitution, but it is not working. I am curious to know, is there any closed form solution for this integral?
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2See Wolfram. Do you want a proof for this? – Bowei Tang Jul 29 '24 at 15:03
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Yes, it will be helpful. – Manoj Kumar Jul 29 '24 at 15:09
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2https://www.integral-calculator.com gives you the complete way of calculation – trula Jul 29 '24 at 15:13
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See https://math.stackexchange.com/a/160533/42969 – Martin R Jul 29 '24 at 15:21
3 Answers
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The shape of the integrand calls for a $\sin^2 t$ substitution.
$$\int\sqrt{\dfrac1x-1}\,dx=\int\sqrt{\frac{1-\sin^2t}{\sin^2t}}2\sin t\cos t\,dt=2\int \cos^2t\,dt=\int(\cos(2t)+1)\,dt.$$
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Let $t=\sqrt{\frac{1}{x}-1}$, then $x=\frac{1}{t^2+1}$ and $dx=\frac{-2t}{(t^2+1)^2}dt$.
\begin{align}\int \sqrt{\frac{1}{x}-1}dx=-2\int\frac{t^2+1-1}{(t^2+1)^2}dt&=-2\int \frac{1}{t^2+1}dt+2\int \frac{1}{(t^2+1)^2}dt\\&=-2\arctan t+2\int \frac{1}{(t^2+1)^2}dt \end{align} For $\int \frac{1}{(t^2+1)^2}dt$, let $t=\tan u$. $$\int \frac{1}{(t^2+1)^2}dt=\int\frac{1}{\sec^4u}\sec^2udu=\int\cos^2udu$$
Bowei Tang
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Substitute $x=\cos^2 u$ as then
$$\sqrt{\dfrac{1}{x}-1}=\sqrt{\sec^2 u -1}=\tan u $$
(motivation - I wanted something which would allow me to simplify the square root) and so
$$\int{\sqrt{\dfrac{1}{x}-1}}dx=\int{\tan u(-2\cos u \sin u )}\ du=\int{-2\sin^2 u}\ du$$
which is
$$\int{\cos 2u - 1}\ du$$ Can you take it from here?
RJG
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