Integral of $$$\frac{1}{\sqrt{a^{2} - u^{2}}}$$$ with respect to $$$u$$$

Find $$$\int \frac{1}{\sqrt{a^{2} - u^{2}}}\, du$$$.

Let $$$u=\sin{\left(v \right)} \left|{a}\right|$$$.

Then $$$du=\left(\sin{\left(v \right)} \left|{a}\right|\right)^{\prime }dv = \cos{\left(v \right)} \left|{a}\right| dv$$$ (steps can be seen »).

Also, it follows that $$$v=\operatorname{asin}{\left(\frac{u}{\left|{a}\right|} \right)}$$$.

So,

$$$\frac{1}{\sqrt{a^{2} - u^{2}}} = \frac{1}{\sqrt{- a^{2} \sin^{2}{\left( v \right)} + a^{2}}}$$$

Use the identity $$$1 - \sin^{2}{\left( v \right)} = \cos^{2}{\left( v \right)}$$$:

$$$\frac{1}{\sqrt{- a^{2} \sin^{2}{\left( v \right)} + a^{2}}}=\frac{1}{\sqrt{1 - \sin^{2}{\left( v \right)}} \left|{a}\right|}=\frac{1}{\sqrt{\cos^{2}{\left( v \right)}} \left|{a}\right|}$$$

Assuming that $$$\cos{\left( v \right)} \ge 0$$$, we obtain the following:

$$$\frac{1}{\sqrt{\cos^{2}{\left( v \right)}} \left|{a}\right|} = \frac{1}{\cos{\left( v \right)} \left|{a}\right|}$$$

Integral becomes

$${\color{red}{\int{\frac{1}{\sqrt{a^{2} - u^{2}}} d u}}} = {\color{red}{\int{1 d v}}}$$

Apply the constant rule $$$\int c\, dv = c v$$$ with $$$c=1$$$:

$${\color{red}{\int{1 d v}}} = {\color{red}{v}}$$

Recall that $$$v=\operatorname{asin}{\left(\frac{u}{\left|{a}\right|} \right)}$$$:

$${\color{red}{v}} = {\color{red}{\operatorname{asin}{\left(\frac{u}{\left|{a}\right|} \right)}}}$$

Therefore,

$$\int{\frac{1}{\sqrt{a^{2} - u^{2}}} d u} = \operatorname{asin}{\left(\frac{u}{\left|{a}\right|} \right)}$$

Add the constant of integration:

$$\int{\frac{1}{\sqrt{a^{2} - u^{2}}} d u} = \operatorname{asin}{\left(\frac{u}{\left|{a}\right|} \right)}+C$$

$$$\int \frac{1}{\sqrt{a^{2} - u^{2}}}\, du = \operatorname{asin}{\left(\frac{u}{\left|{a}\right|} \right)} + C$$$A