Evaluating ${\displaystyle\int_0^{15\pi}\sqrt{\left(R+\frac{h\theta}{2\pi}\right)^2+\frac{h^2}{4\pi^2}}\,\,\,\mathrm d\theta}$

Question

How would one go about evaluating the following integral? $$\int_0^{15\pi}\sqrt{\left(R+\frac{h\theta}{2\pi}\right)^2+\frac{h^2}{4\pi^2}}\,\,\,\mathrm d\theta$$

I tried expanding the $\left(R+\frac{h\theta}{2\pi}\right)^2$ to get $$\int_0^{15\pi}\sqrt{R^2+\frac{Rh\theta}{\pi}+\frac{h^2\theta^2}{4\pi^2}+\frac{h^2}{4\pi^2}}\,\,\,\mathrm d\theta$$ then factoring the $\frac{h^2}{4\pi^2}$ out $$\int_0^{15\pi}\frac{h}{2\pi}\sqrt{\frac{4\pi^2R^2}{h^2}+\frac{4\pi R\theta}{\pi}+\theta^2+1}\,\,\,\mathrm d\theta$$ and factorising by the difference of two squares $$\frac{h}{2\pi}\int_0^{15\pi}\sqrt{\left(\theta+\frac{2\pi R}{h}\right)^2+1}\,\,\,\mathrm d\theta$$ then using the substitution $\tan(u) = \theta + \frac{2\pi R}{h}$ to get $$\frac{h}{2\pi}\int_{\theta=0}^{\theta=15\pi}\sqrt{\tan^2(u)+1}\,\,\,\mathrm d\theta\\ = \frac{h}{2\pi}\int_{\theta=0}^{\theta=15\pi}\sec(u)\,\,\,\mathrm d u= \frac{h}{2\pi}\bigg[\ln\mid\sec(u)+\tan(u)\mid\bigg]_\frac{2\pi R}{h}^{\frac{2\pi R}{h}+15\pi}$$However, this result is very different to that given by Wolfram|Alpha, not just symbolically (one could have used $\cosh^{-1}(\sec(u))$ instead of the $\ln$ I used, for example), but the result I get by substituting in values for $R$ and $h$ is also wrong. Could somebody kindly point me in the right direction?

Answer 1

The mistake is: You didn't substitute $d\theta = \sec^2u \ du$.

To integrate $\int \sec^3 u du = \int\frac{\cos u}{(1-\sin^2u)^2} du$

Put $z = \sin u$ and then use partial fraction method.

$$\int\frac{dz}{(1-z^2)^2} = \frac{1}{4}\int \frac{1}{1-x} + \frac{1}{(1-x)^2} + \frac{1}{1+x} + \frac{1}{(1+x)^2} dx$$

Now, take it from here.

Answer 2

I would first factor out $h/(2\pi)$; i.e. the integrand would become $$\left(\left(R + \frac{h\theta}{2\pi}\right)^{\!2} + \frac{h^2}{4\pi^2}\right)^{\!1/2} = \frac{h}{2\pi} \left(\left(R' + \theta\right)^2 + 1\right)^{\!1/2},$$ where $$R' = \frac{2\pi R}{h}.$$ Now the appropriate substitution is evident: let $$\tan \varphi = R' + \theta, \quad \sec^2 \varphi \, d\varphi = d\theta.$$ This yields $$\int_{\theta = 0}^{15\pi} \left(\left(R + \frac{h\theta}{2\pi}\right)^{\!2} + \frac{h^2}{4\pi^2}\right)^{\!1/2} \, d\theta = \frac{h}{2\pi} \int_{\varphi = \tan^{-1} R'}^{\tan^{-1} R' + 15\pi} \sec^3 \varphi \, d\varphi,$$ which is handled by the usual means.

Answer 3

Instead of $\tan(u) = \theta + \frac{2\pi R}{h},$ try $$w = \theta + \frac{2\pi R}{h}$$ which gives some $$ E_1 \int_A^B \; \sqrt {w^2 + 1} \; dw. $$ Then $w = \sinh t$ and $dw = \cosh t dt$ uses no absolute value signs, just $$ E_1 \int_C^D \; \cosh^2 t \; dt. $$ I believe next is $$ \cosh 2t = 2 \cosh^2 t - 1, $$ need to check that. https://en.wikipedia.org/wiki/Hyperbolic_function#Sums_of_arguments