Compute $\int\frac{du}{\sqrt{a^2-u^2}}$

Question

Please, can you explain why isn't the $$\int\frac{du}{\sqrt{a^2-u^2}} = \int\frac{du}{a\sqrt{1-(u/a)^2}}=\frac1a\sin^{-1}\bigg(\frac{u}{a}\bigg)+c$$ if $a > 0$ is a positive constant.

Answer 1

Because when you do a change of variables, it affects the differential element as well: if $v=\frac ua$ then $dv=\frac 1a du$.

As you can see also when you compute the derivative of your right-hand side, a second denominator comes out and you end up with $\frac 1{a^2}$.

Answer 2

What is true is that

$$\int\frac{du}{\sqrt{a^2-u^2}} = \int\frac{d(u/a)}{\sqrt{1-(u/a)^2}}= \sin^{-1}(u/a)+c$$

Answer 3

We have $$\int\frac{du}{\sqrt{a^2-u^2}}.$$

Let $u=a\sin\theta\implies\theta=\arcsin\bigg(\dfrac{u}{a}\bigg)$. Then $du=a\cos\theta\ d\theta$ and we have

$$\int\frac{du}{\sqrt{a^2-u^2}}=\int\frac{a\cos\theta\ d\theta}{\sqrt{a^2-a^2\sin^2\theta}}=\int\frac{a\cos\theta\ d\theta}{a\sqrt{1-\sin^2\theta}}=\int\frac{\cos\theta\ d\theta}{\sqrt{\cos^2\theta}}=\int d\theta$$ $$=\theta=\boxed{\arcsin\bigg(\frac{u}{a}\bigg)+C}.$$

Answer 4

Let $v = \dfrac ua \implies u = a\cdot v$, and $\mathrm du = a\cdot\mathrm dv$.

Substitute in original expression,

$$\require{cancel}\begin{align}\int\dfrac{\mathrm du}{\sqrt{a^2 - u^2}} &\equiv \int\dfrac{a\cdot\mathrm dv}{\sqrt{a^2 - a^2v^2}}\\ &= \int\dfrac{a\cdot\mathrm dv}{a\sqrt{1 - v^2}}\\ &= \int\dfrac{\cancel{a}\cdot\mathrm dv}{\cancel{a}\sqrt{1 - v^2}}\\ &= \int\dfrac{\mathrm dv}{\sqrt{1 - v^2}} = \sin^{-1}\left(v\right) + C\end{align}$$

Reverse substitution, $$\sin^{-1}\left(v\right) + C \equiv \sin^{-1}\left(\dfrac ua\right) + C$$

What you are missing in your calculation is that $$\dfrac {\mathrm d}{\mathrm du}\sin^{-1}\left(\dfrac ua\right) = \dfrac 1a\cdot\dfrac1{\sqrt{1 - \left(u/a\right)^2}}$$

and, therefore, the $1/a$ is consumed as part of the integral.