Mathematics

Solve : $$\underset{0}{\overset{1}{\int}} \sqrt{9 - 4x^2} dx$$


SOLUTION
$$\int \sqrt{9-4x^2}dx$$

$$x=\dfrac{3}{2}\sin u\rightarrow dx=\dfrac{3}{2}\cos u du$$

$$=\int \left ( \sqrt{9-4\frac{9}{4}\sin^2 u} \right )\frac{3}{2}\cos u du$$

$$=\int \sqrt{9(1-\sin^2 u)}\frac{3}{2}\cos u du$$

$$=\int (3\cos u) \frac{3}{2}\cos u du$$

$$=\dfrac{9}{2}\int \cos ^2 u du$$

$$=\dfrac{9}{2}\int\left ( \frac{1+\cos 2u}{2} \right )du$$

$$=\dfrac{9}{2}\cdot \dfrac{1}{2}\left ( u+\dfrac{1}{2}\sin u \right )$$

$$u=\sin^{-1}\left ( \dfrac{2}{3}x \right )$$

$$\int \sqrt{9-4x^2}dx=\dfrac{9}{2}\cdot \dfrac{1}{2}\left ( \sin^{-1}\left ( \dfrac{2}{3}x \right )+\dfrac{1}{2}\sin\left ( \sin^{-1}\left ( \dfrac{2}{3}x \right ) \right )\right )+C$$

$$\int_{0}^{1}\sqrt{9-4x^2}dx=\dfrac{9}{4}\sin^{-1}\left ( \dfrac{2}{3} \right )+\dfrac{\sqrt{5}}{2}$$
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Subjective Medium Published on 17th 09, 2020
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