Question

Explain why, if $x=a\sin u$, then $\sqrt{a^2-x^2}=a\cos u$. Your explanation should include a discussion of domains and absolute values.

Short answer

Ans:

$$\begin{gathered} \sqrt{a^2-x^2}=a\cos u.....[x=a\sin u] \\ =\sqrt{a^2-a^2{\sin }^{2}u} \\ =a\cos u \end{gathered}$$

Step-by-step solution

Step 1. Given information:

$\sqrt{a^2-x^2}=a\cos u$

$x=a\sin u$

Step 2. Solving the trigonometric substitution:

If $x=a\sin u$, then $x$must be between $-a$and $a$. Therefore the substitution $x=a\sin u$makes sense only for integrands whose domain is contained in [-a, a] .

The domain of $\sqrt{a^2-x^2}$is precisely $[-a,a]$, so the integral involving $\sqrt{a^2-x^2}$are well suited for trigonometric substitution with $x=a\sin u$.

Using the Pythagorean identity $a^2{\sin }^{2}u+a^2{\cos }^{2}u=a^2$,

$$\begin{gathered} \sqrt{a^2-x^2} \\ =\sqrt{a^2-a^2{\sin }^{2}u} \\ =\sqrt{a^2{\cos }^{2}u} \\ =\left|a\cos u\right|[A.\sqrt{A^2}=\left|A\right|] \\ =a\cos u \end{gathered}$$