Question

The integral $ \int \sqrt{16-9x^2} $ $dx$ equals

• A. $ \frac{x}{2}\sqrt{16-9x^{2}}+\frac{8}{3}sin^{-1}\left(\frac{3x}{4}\right)+C $
• B. $ \frac{3x}{2}\sqrt{16-9x^{2}}+ 16 sin^{-1} \left(\frac{3x}{4}\right)+C $
• C. $ \frac{\pi}{2} sin^{-1} (\frac {3x}{4}) + \frac{9x}{2} +C $
• D. $None\, of\, the\, above$

Answer 1

Answer: (A)

Let $ I = \int \sqrt{16-9x^2} \,dx$
$ = \int \sqrt{9(\frac{16}{9} - x^2)} \,dx$
$ = 3 \int \sqrt{(\frac{4}{3})^2 - x^2} \,dx$
$= 3\left[\frac{x}{2}\times \frac{\sqrt{16-9x^{2}}}{3}+\frac{\left(\frac{4}{3}\right)^{2}}{2} \sin^{-1}\left(\frac{x}{\frac{4}{3}}\right)\right] + C$
$ = 3\left[\frac{x}{2}\times \frac{\sqrt{16-9x^{2}}}{3} + \frac{16}{9\times2}\sin^{-1}\left(\frac{3x}{4}\right)\right]+C$
$ = \frac{x}{2}\sqrt{16-9x^{2}} + \frac{8}{3} \sin^{-1}\left(\frac{3x}{4}\right)+ C$