Cosecant Inverse Integarl

First, let's take the indefinite integral using integration by parts.

$\displaystyle I =\int \csc^{-1} (x) \,\mathrm{d}x$

Let $u= \csc^{-1} (x)$ , and $\,\mathrm{d}v=\mathrm{d}x$ , so $\,\mathrm{d}u=-\dfrac{\mathrm{d}x}{x\sqrt{x^2-1}}$ (for $x>0$ ), and $\,v=x$ . Then, $\displaystyle I = uv-\int v\,\mathrm{d}u = x\csc^{-1}(x) + \int \frac{\mathrm{d}x}{\sqrt{x^2-1}}$

Now if we make the trig. substitution $\,x=\sec \theta\,\,\,\Rightarrow\,\,\,\mathrm{d}x=\sec\theta\tan\theta$ ,

$\begin{aligned} \displaystyle \rightarrow \int \frac{\mathrm{d}x}{\sqrt{x^2-1}}=\int \sec \theta\,\mathrm{d} \theta =\ln\left(\sec\theta+\tan\theta\right) &= \ln\left(\sec\theta + \sqrt{\sec^2\theta -1}\right) \\ &=\ln\left(x+\sqrt{x^2-1}\right) \end{aligned}$

(for $x>0$ , ignoring the constant of integration)

So we have $\displaystyle I=x\csc^{-1}(x) + \ln\left(x+\sqrt{x^2-1}\right)$

Now, our definite integral is $I(99)-I(1)$ .

$I(1)=\csc^{-1}(1) = \sin^{-1}(1)=\dfrac{\pi}{2}.$ We can approximate $\dfrac{\pi}{2}$ as $1.57$ .

Here's the fun part, approximating our way to

$I(99)=99\csc^{-1}(99) + \ln\left(99+\sqrt{99^2-1}\right)$

For the first term, since $\csc^{-1}(x)=\sin^{-1}\left(\dfrac{1}{x}\right)$ , we want $99\sin^{-1}\left(\dfrac{1}{99}\right)$ .

Now we can use the small-angle approximation $\,\sin x \approx x\,$ (since $\frac{1}{99}$ is close enough to $\,0$ ), so that $\sin^{-1}\left(\dfrac{1}{x}\right) \approx \dfrac{1}{x}$ . Then $\,99\sin^{-1}\left(\dfrac{1}{99}\right)=99\csc^{-1}(99)\,$ is pretty close to $1$ .

For the second term, after working $99^2$ out, we want to find $\ln(99+70\sqrt{2})$ .

Approximating $\sqrt{2}$ as $1.414$ , $70\times 1.414=98.98$ .

In order to find $\ln(99+98.98)=\ln(197.98)$ , we can split this into

$\ln(1.9798)+\ln(10^2)=\ln(1.9798)+2\ln(10)$ .

I got pretty lucky like Richard Feynman this time, because I don't know by heart the natural logs of anything besides $2$ and $10$ , and I think I even learned $\ln10$ from Feynman's Surely You're Joking, Mr. Feynman! .

Anyway, we can approximate $\ln 1.9798$ as $\ln 2$ which is approximately $0.69$ , and we can approximate $\ln 10$ as $2.3$ .

Notice that this method of taking $\,\ln(x \cdot 10^n)=\ln x + n\ln 10\,$ allows us to decently approximate the natural logs of really large or small numbers, as long as you know the nat. logs of the first 10 integers; pretty cool!

Back to the problem, we then have that $\ln(197.98) \approx \ln 2 + 2\ln10 \approx 0.69+4.6 =5.29$ .

So we have $I(99)\approx 1+5.29=6.29$ .

Finally, $\displaystyle \int_1^{99} \csc^{-1} (x) \,\mathrm{d}x = I(99)-I(1) \approx 6.29 - 1.57 = \boxed{4.72}$ .

$\begin{aligned} I & = \int_1^{99} {\color{#3D99F6}1} \cdot {\color{#D61F06} \csc^{-1} x} \ dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = {\color{#3D99F6}x}{\color{#D61F06} \csc^{-1} x}\ \bigg|_1^{99} {\color{#D61F06}+} \int_1^{99} \frac {\color{#3D99F6}x}{\color{#D61F06}x\sqrt{x^2-1}} \ dx & \small \color{#3D99F6} \text{Note that }\frac d{dx} \csc^{-1} x = - \frac 1{x\sqrt{x^2-1}} \\ & = 99\csc^{-1}99 - \frac \pi 2 + \int_1^{99} \frac 1{\sqrt{x^2-1}} \ dx & \small \color{#3D99F6} \text{Let }x = \sec \theta \implies dx = \sec \theta \tan \theta \ d \theta \\ & = 99\csc^{-1}99 - \frac \pi 2 + \int_0^{\sec^{-1} 99} \frac {\sec \theta \tan \theta}{\sqrt{\sec^2 \theta -1}} \ d\theta \\ & = 99\csc^{-1}99 - \frac \pi 2 + \int_0^{\sec^{-1} 99} \sec \theta \ d\theta \\ & = 99\csc^{-1}99 - \frac \pi 2 + \ln(\tan \theta + \sec \theta) \ \bigg|_0^{\sec^{-1} 99} \\ & = 99\csc^{-1}99 - \frac \pi 2 + \ln \left(70\sqrt 2 + 99 \right) - \ln \left(0+1 \right) \\ & = 99\csc^{-1}99 - \frac \pi 2 + \ln \left(70\sqrt 2 + 99 \right) \\ & \approx \boxed{4.717} \end{aligned}$