Question

${\int }_0^a\sqrt{a^2-x^2}dx=$

• A. $\pi a^2$
• B. $\frac{\pi a^2}{2}$
• C. $\frac{\pi a^2}{3}$
• D. $\frac{\pi a^2}{4}$

Answer 1

The correct option is D

$\frac{\pi a^2}{4}$

Explanation for the correct option:

Let $I={\int }_0^a\sqrt{a^2-x^2}dx$

Let us consider $x=a\sin \left(\theta \right)$

$\Rightarrow dx=a\cos \left(\theta \right)$

$$\begin{gathered} I={\int }_0^a\sqrt{a^2-a^2{\sin }^2\left(\theta \right)}a\cos \left(\theta \right)d\theta \\ ={\int }_0^a\sqrt{a^2(1-{\sin }^2\left(\theta \right)}a\cos \left(\theta \right)d\theta \\ ={\int }_0^a\sqrt{a^2{\cos }^2\left(\theta \right)}a\cos \left(\theta \right)d\theta \\ ={\int }_0^{a}a\cos \left(\theta \right)a\cos \left(\theta \right)d\theta \\ =a^2{\int }_0^a{\cos }^2\left(\theta \right)d\theta \end{gathered}$$

Now we know that, ${\cos }^2\left(\theta \right)=\frac{1+\cos \left(2\theta \right)}{2}$.

Substituting this value in the above integral we get,

$$\begin{gathered} I=a^2{\int }_0^a\left(\frac{1+\cos \left(2\theta \right)}{2}\right)d\theta \\ =a^2{\left[\frac{\theta }{2}+\frac{\sin \left(2\theta \right)}{4}\right]}_0^a+C\left[\because \int \cos \left(2\theta \right)=\sin \left(2\theta \right)\right] \\ =a^2{\left[\frac{{\sin }^{-1}\left({\displaystyle \frac{x}{a}}\right)}{2}+\frac{2\sin \left(\theta \right)\cos \left(\theta \right)}{4}\right]}_0^a+C\left[x=a\sin \left(\theta \right)\Rightarrow \theta ={\sin }^{-1}\left(\frac{x}{a}\right)\text{and}\sin \left(2\theta \right)=2\sin \left(\theta \right)\cos \left(\theta \right)\right] \\ =a^2{\left[\frac{{\sin }^{-1}\left({\displaystyle \frac{x}{a}}\right)}{2}+\frac{2{\displaystyle \frac{x}{a}}{\displaystyle \frac{\sqrt{a^2-x^2}}{a}}}{4}\right]}_0^a+C \\ =a^2\left[\frac{\pi }{4}+0-0-0\right]+C \\ =a^2\left[\frac{\pi }{4}\right]+C \\ I=\frac{\pi a^2}{4}+C \end{gathered}$$

Hence, option D is the correct answer.