Question

Evaluate the integral
${\int }_{\frac{a}{2}}^a\frac{1}{\sqrt{a^2-x^2}}dx$

• A. $\frac{\pi }{2}$
• B. $\pi a$
• C. $\pi -1$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is D $\frac{\pi }{3}$

${\int }_a^{\frac{a}{2}}\frac{1}{\sqrt{a^2-x^2}}\cdot dx$

$\int \frac{1}{\sqrt{a^2-x^2}}\cdot dx=\frac{1}{a}\int \frac{1}{\sqrt{1-{\left(\frac{x}{a}\right)}^2}}\cdot dx$

$=\int \frac{d\left(\frac{x}{a}\right)}{\sqrt{1-{\left(\frac{x}{a}\right)}^2}}$

$={\sin }^{-1}\left(\frac{x}{a}\right)+c$

${\int }_{\frac{a}{2}}^a\frac{1}{\sqrt{a^2-x^2}}\cdot ={\left[{\sin }^{-1}\right(\frac{x}{a})+c]}_{\frac{a}{2}}^a$

$=\left({\sin }^{-1}\right(\frac{a}{a})+c)-\left({\sin }^{-1}\right(\frac{\frac{a}{2}}{a})+c)$

$=\left[\right(\sin \left(1\right)+c)-({\sin }^{-1}\left(\frac{1}{2}\right)+c\left)\right]$

$=\frac{\pi }{2}-\frac{\pi }{6}$

$=\frac{\pi }{3}$