Question

\[\int\sqrt{a^2 - x^2}\text{  dx }\]

Answer 1

\[\text{ Let I } = \int\sqrt{a^2 - x^2} \text{  dx }\]
\[ = \int {1  _{II} \cdot} \sqrt{a^2 { _I} - x^2} dx\]
\[ = \sqrt{a^2 - x^2}_{} \int1 \text{  dx }- \int\left( \frac{d}{dx}\left( \sqrt{a^2 - x^2} \right)\int1\text{  dx } \right)dx\]
\[ = \sqrt{a^2 - x^2} \cdot x + \int\frac{1 \times 2x}{2 \sqrt{a^2 - x^2}} \cdot x\text{  dx }\]
\[ = \sqrt{a^2 - x^2} \cdot x + \int\left( \frac{x^2 - a^2 + a^2}{\sqrt{a^2 - x^2}} \right) dx\]
\[ = x\sqrt{a^2 - x^2} - \int\sqrt{a^2 - x^2} dx + a^2 \int\frac{1}{\sqrt{a^2 - x^2}}dx\]
\[ = x\sqrt{a^2 - x^2} - I + a^2 \int\frac{1}{\sqrt{a^2 - x^2}}dx\]
\[ \therefore 2I = x\sqrt{a^2 - x^2} + a^2 \int\frac{1}{\sqrt{a^2 - x^2}}dx\]
\[ \Rightarrow I = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{- 1} \left( \frac{x}{a} \right) + C\]