Question

\[\int\cos x \sqrt{4 - \sin^2 x}\text{ dx}\]

Answer 1

\[\text{ Let I} = \int \cos x \sqrt{4 - \sin^2 x}\text{ dx}\]
\[\text{ Putting  sin x} = t\]
\[ \Rightarrow \cos\ x\ dx = \text{ dt }\]
\[\int \sqrt{2^2 - t^2} \text{ dt }\]
\[ = \frac{t}{2}\sqrt{2^2 - t^2} + \frac{2^2}{2} \text{ sin}^{- 1} \left( \frac{t}{2} \right) + C \left[ \because \int\sqrt{a^2 - x^2} \text{ dx } = \frac{1}{2}x\sqrt{a^2 - x^2} + \frac{1}{2} a^2 \sin^{- 1} \left( \frac{x}{a} \right) + C \right]\]
\[ = \frac{\sin x}{2} \sqrt{4 - \sin^2 x} + 2 \sin^{- 1} \left( \frac{\sin x}{2} \right) + C \left[ \because t = \sin x \right]\]