How to find reduction formula for $I_n=\int\frac{1}{x^n \sqrt{x^2-1}}dx$

Question

I have tried manipulating the integral but couldn't come up with an answer that matched textbook answer. I just need a hint. Thanks in advance. $\int\frac{dx}{x^n \sqrt{x^2-1}}$

This is what I tried: Let $I_{n-2}=\int\frac{dx}{x^{n-2} \sqrt{x^2-1}}$ Then I integrated by parts.

The answer given in textbook is

$$(n-1)I=\frac{\sqrt{x^2-1}}{x^{n-1}} +(n-2)I_{n-2}$$

Answer 1

Reduce the integral as follows

\begin{align} I_n= &\int\frac{1}{x^n \sqrt{x^2-1}}dx \\ =& \int\frac{(1-x^2)+x^2}{x^n \sqrt{x^2-1}}dx =-\int \frac{\sqrt{x^2-1}}{x^n }dx + I_{n-2} \\ =&\ \frac1{n-1}\int \sqrt{x^2-1} \>d\left(\frac{1}{x^{n-1} }\right)+ I_{n-2} \\ \overset{ibp}=&\ \frac1{n-1}\frac{\sqrt{x^2-1}}{x^{n-1} }+\frac{n-2}{n-1}I_{n-2} \end{align}

Answer 2

The substitution makes the integral $$\int\cos^{n-1}(θ)\,dθ$$ if I am right and the absolute value is ignored. The reduction formula is $\cos^{n-2}θ\sinθ+(n-2)(\int \cos^{n-3}θ\,dθ-\int\cos^{n-1}\,dθ)$

This is after a bit of integration by parts of $\cos^{n-2}θ$ and $\cosθ$, pythagorean identity and basic integral properties. Also,remember that $x=\secθ$ just as @Tavish said.

This will not look like the textbook form, but using tan(arcsecθ)=xsin(arcsecθ)=x$\sqrt{1-\frac {1}{x^2}}$ and substitution of the change of variable got me dx= $x^2\sqrt{1-\frac {1}{x^2}}$dθ=secθtanθdθ.

Finally we get your original integral after some simplification is: $x^{2-n}\sqrt{1-\frac{1}{x^2}}+(n-2)[\int \frac{x^{3-n}-x^{1-n}}{|x|\sqrt{x^2-1}}dx]$= $\frac{\sqrt{x^2-1}}{|x|x^{n-2}}+(n-2)[\int \frac{x^{3-n}}{|x|\sqrt{x^2-1}}dx-\int\frac{x^{1-n}}{|x|\sqrt{x^2-1}}dx]$

You can put it inti $I_{n-N}$ form if needed.

Correct me and give me feedback please!

Answer 3

I would do the hyperbolic substitution $\:x=\cosh t,\enspace t\ge 0$, $\:\mathrm d x=\sinh t\,\mathrm d t$, so that the integral becomes, since $\:\cosh^2t-\sinh^2t=1$, $$\int\frac{\sinh t\,\mathrm dt}{\sinh t\cosh^nt}=\int\frac{\mathrm dt}{\cosh^nt},$$ that we'll denote $I_n$.

Now, to obtain the recurrence relation, we'll use integration by parts for $I_n$: setting \begin{cases}u=\dfrac 1{\cosh^{n-2}t}, &\quad\mathrm du=-\dfrac{(n-2)}{\cosh^{n-1}t}\sinh t\,\mathrm dt, \\[1ex] \mathrm dv=\dfrac{\mathrm dt}{\cosh^2t},&\quad v=\tanh t. \end{cases} we obtain $$I_n=\frac{\tanh t}{\cosh^{n-2}t}+(n-2)\int\frac{\tanh t\,\mathrm dt}{\cosh^{n-1}t}=\frac{\sinh t}{\cosh^{n-1}t}+(n-2)\int\frac{\sinh^2 t\,\mathrm dt}{\cosh^{n}t}$$

Can you proceed?

Answer 4

Hint:

With $$I_n:=\int\frac{dx}{x^n\sqrt{x^2-1}},$$

$$I_{n-2}-I_{n}=\int\frac{x^2-1}{x^{n}\sqrt{x^2-1}}dx=\int\frac{\sqrt{x^2-1}}{x^{n}}dx.$$

Then by parts, this is

$$-\frac1{n-1}\frac{\sqrt{x^2-1}}{x^{n-1}}+\frac1{n-1}I_{n-2}.$$