Evaluate the integral using an appropriate trigonometric substitution.
$$\int _ { 0 } ^ { \sqrt { 2 } } \frac { x ^ { 2 } } { \sqrt { 4 - x ^ { 2 } } } d x$$
A) $\pi - 1$
B) $\pi$
C) $\frac { 2 \pi } { 3 } - \frac { \sqrt { 3 } } { 2 }$
D) $\frac { \pi } { 2 } - 1$

Question

Quizplus · Accepted Answer

Using the substitution $x = 2\sin\theta$, we have $dx = 2\cos\theta\,d\theta$ and $\sqrt{4-x^2} = \sqrt{4-4\sin^2\theta} = 2\cos\theta$. Hence, the integral becomes
$$\int_0^{\frac{\pi}{4}} \frac{(2\sin\theta)^2}{2\cos\theta} (2\cos\theta)\,d\theta = 4\int_0^{\frac{\pi}{4}} \sin^2\theta\,d\theta.$$
Using the identity $\sin^2\theta = \frac{1}{2}(1-\cos2\theta)$, we get
$$\int_0^{\frac{\pi}{4}} \sin^2\theta\,d\theta = \frac{1}{2}\int_0^{\frac{\pi}{4}} (1-\cos2\theta)\,d\theta = \frac{1}{2}\left(\theta - \frac{1}{2}\sin2\theta \right)\bigg|_0^{\frac{\pi}{4}} = \frac{\pi}{8} - \frac{1}{4}.$$
Therefore, the answer is $\frac{\pi}{2} - 1$, which corresponds to choice $\boxed{\text{(D)}}$.

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Evaluate the Integral Using an Appropriate Trigonometric Substitution $\int _ { 0 } ^ { \sqrt { 2 } } \frac { x ^ { 2 } } { \sqrt { 4 - x ^ { 2 } } } d x$