Question:
The integral $ \int \sqrt{16-9x^2} $ $dx$ equals
The integral $ \int \sqrt{16-9x^2} $ $dx$ equals
Updated On: Jun 14, 2022
- $ \frac{x}{2}\sqrt{16-9x^{2}}+\frac{8}{3}sin^{-1}\left(\frac{3x}{4}\right)+C $
- $ \frac{3x}{2}\sqrt{16-9x^{2}}+ 16 sin^{-1} \left(\frac{3x}{4}\right)+C $
- $ \frac{\pi}{2} sin^{-1} (\frac {3x}{4}) + \frac{9x}{2} +C $
- $None\, of\, the\, above$
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The Correct Option is A
Solution and Explanation
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$
$ = \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$
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Concepts Used:
Integrals of Some Particular Functions
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
