Question

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

• A. $\pi$
• B. $\frac{\pi }{3}$
• C. $-\pi$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (D)

$\frac{\pi }{4}$

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

Substituting $x=a\sin t\Rightarrow dx=a\cos tdt$

$I={\int }_0^{\frac{\pi }{2}}\frac{\cos tdt}{\sin t+\cos t}$ ...(1)

Using property ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$I={\int }_0^{\frac{\pi }{2}}\frac{\sin tdt}{\cos t+\sin t}$ ...(2)

Adding (1) and (2), we get

$2I={\int }_0^{\frac{\pi }{2}}dt=\frac{\pi }{2}\Rightarrow I=\frac{\pi }{4}$