9 4. evaluate the integral ∫₀^√2 1/(4 - x²)^(3/2) dx.

Explanation:

Step1: Use trig - substitution

Let (x = 2\sin\theta), then (dx=2\cos\theta d\theta). When (x = 0), (\theta=0); when (x=\sqrt{2}), (\sin\theta=\frac{\sqrt{2}}{2}), so (\theta=\frac{\pi}{4}). Also, (4 - x^{2}=4 - 4\sin^{2}\theta = 4\cos^{2}\theta).

Step2: Rewrite the integral

The integral (\int_{0}^{\sqrt{2}}\frac{1}{(4 - x^{2})^{\frac{3}{2}}}dx=\int_{0}^{\frac{\pi}{4}}\frac{2\cos\theta}{(4\cos^{2}\theta)^{\frac{3}{2}}}d\theta=\int_{0}^{\frac{\pi}{4}}\frac{2\cos\theta}{8\cos^{3}\theta}d\theta=\frac{1}{4}\int_{0}^{\frac{\pi}{4}}\sec^{2}\theta d\theta).

Step3: Integrate (\sec^{2}\theta)

We know that (\int\sec^{2}\theta d\theta=\tan\theta + C). So (\frac{1}{4}\int_{0}^{\frac{\pi}{4}}\sec^{2}\theta d\theta=\frac{1}{4}[\tan\theta]_{0}^{\frac{\pi}{4}}).

Step4: Evaluate the definite - integral

(\frac{1}{4}(\tan\frac{\pi}{4}-\tan0)=\frac{1}{4}(1 - 0)=\frac{1}{4}).

Answer:

(\frac{1}{4})