Question

Integrate the function w..r. to x : $\sqrt{4-9x^{2}}$

Answer 1

Solution:

Given equation can be converted into the standard equation as shown below

$\int \sqrt{a^{2}-x^{2}} dx= \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\:sin^{-1}\:(\frac{x}{a})+C$

By taking out 9 common we can write the given expression as

$\int \sqrt{(2)^{2}-(3x)^{2}} dx\\\\=3\int\sqrt{(\frac{2}{3})^{2}-x^{2}} dx$

Now put this term into the formula

$\int \sqrt{(\frac{2}{3})^{2}-x^{2}} dx= \frac{x}{2} \sqrt{(\frac{2}{3})^{2}-x^{2}}+\frac{(\frac{2}{3})^{2}}{2}\:sin^{-1}\:(\frac{x}{\frac{2}{3}})+C\\\\= \frac{x}{2(3)}\sqrt{(4-9x^{2}}+\frac{2}{9}\:sin^{-1}\:(\frac{3x}{2})+C\\\\3\int\sqrt{(\frac{2}{3})^{2}-x^{2}} dx=\frac{x}{2}\sqrt{(4-9x^{2}}+\frac{2}{3}\:sin^{-1}\:(\frac{3x}{2})+C$

So
$\int \sqrt{4-9x^{2}} dx=\frac{x}{2}\sqrt{4-9x^{2}}+\frac{2}{3}\:sin^{-1}\:(\frac{3x}{2})+C$