Question

Fill in the blanks to complete each of the following theorem statements:

If $a$is a positive real number, and if $x\in \_\_\_\_\_$and $u\in \_\_\_\_\_$, then the substitution $x=a\sin u$gives us $\sqrt{a^2-x^2}=\_\_\_\_\_$.

Short answer

If $a$is a positive real number, and if $x\in \left[-a,a\right]$and $u\in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$ , then the substitution $x=a\sin u$ gives us $\sqrt{a^2-x^2}=a\cos u$.

Step-by-step solution

Step 1. Given information  

If $a$is a positive real number, and if role="math" localid="1653537491868" $x\in \_\_\_\_\_$and $u\in \_\_\_\_\_$, then the substitution $x=a\sin u$ gives us $\sqrt{a^2-x^2}=\_\_\_\_\_$.

Step 2. Filling in the blanks to complete the theorem statements 

If $a$is a positive real number, and if $x\in \left[-a,a\right]$and $u\in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$, then the substitution $x=a\sin u$gives us $\sqrt{a^2-x^2}=a\cos u$.

By substituting for $x$and simplifying, we get the above answer valid under the conditions mentioned. That is, $\sqrt{a^2-x^2}=\sqrt{a^2-{\left(a\sin u\right)}^2}$

$$\begin{gathered} =\sqrt{a^2-a^2{\sin }^{2}u} \\ =\sqrt{a^2(1-{\sin }^{2}u)} \\ =\sqrt{a^2{\cos }^{2}u} \\ =a\cos u \end{gathered}$$