MHB Solve Integration Problem - Get Help Now!

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To solve the integral $$\int \frac{ds}{\sqrt{s^2-0.01}}$$, a substitution method is suggested. Using the substitution $s = r \sec(x)$ simplifies the integrand due to the identity $\sec^2(x) - 1 = \tan^2(x)$. Another effective substitution is $s = r \cosh(t)$, which transforms the differential and may simplify the integral further. Both methods aim to facilitate the integration process by reducing complexity in the integrand. Exploring these substitutions should yield a clearer path to the solution.
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Hello can you help solve this problem

$$\int_{}^{}\frac{ds}{\sqrt{s^2-0.01}}$$

I tried using method of substitution but I still could not find a good cancellation.
Please tell me what to do. Thanks!
 
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Integrals of that form can be generalized to
$$\int \frac{ds}{\sqrt{s^2-r^2}}$$

Using the substitution $s=r \sec\left({x}\right)$ simplifies the integrand because of the property $\sec^2\left({x}\right)-1=\tan^2\left({x}\right)$.

Trying this, what do you get?
 
The substitution $\displaystyle \begin{align*} s = r \cosh{(t)} \implies \mathrm{d}s = r\sinh{(t)}\,\mathrm{d}t \end{align*}$ may lead to a simpler integrand.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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