Question

${\int }_0^1\frac{d\mathrm{x}}{\mathrm{x}+\sqrt{1-{\mathrm{x}}^2}}=$

• A. $\frac{\mathrm{\pi }}{3}$
• B. $\frac{\mathrm{\pi }}{2}$
• C. $\frac{1}{2}$
• D. $\frac{\mathrm{\pi }}{4}$

Answer 1

The correct option is D

$\frac{\mathrm{\pi }}{4}$

Explanation for the correct option:

Evaluating ${\int }_0^1\frac{d\mathrm{x}}{\mathrm{x}+\sqrt{1-{\mathrm{x}}^2}}$:

Let $\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{d\mathrm{x}}{\mathrm{x}+\sqrt{1-{\mathrm{x}}^2}}$

Let's consider $\mathrm{x}=\sin \left(\mathrm{\theta }\right)$.

Differentiating it with respect to $\mathrm{x}$ we get

$d\mathrm{x}=\cos \left(\mathrm{\theta }\right)d\mathrm{\theta }$

Substituting these values in the above integral we get,

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

We know that ${\int }_a^{b}f\left(x\right)=f\left(a+b-x\right)dx$

Hence, $0+\frac{\mathrm{\pi }}{2}-\mathrm{\theta }=\frac{\mathrm{\pi }}{2}-\mathrm{\theta }$ [ where$a=0,b=\frac{\mathrm{\pi }}{2},x=\theta$]

Therefore,

$$\begin{gathered} \mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-\mathrm{\theta }\right)}{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-\mathrm{\theta }\right)+\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-\mathrm{\theta }\right)}d\mathrm{\theta } \\ \mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin \left(\mathrm{\theta }\right)}{\cos \left(\mathrm{\theta }\right)+\sin \left(\mathrm{\theta }\right)}d\mathrm{\theta }\dots \dots \left(2\right) \end{gathered}$$

Adding equations $\left(1\right)$ and $\left(2\right)$ we get,

$\begin{array}{rcl}\text{I+I}& =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\cos \left(\theta \right)}{\sin \left(\theta \right)+\cos \left(\theta \right)}d\theta +{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin \left(\theta \right)}{\sin \left(\theta \right)+\cos \left(\theta \right)}d\theta \\ & \Rightarrow & 2\text{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\cos \left(\theta \right)+\sin \left(\theta \right)}{\sin \left(\theta \right)+\cos \left(\theta \right)}d\theta \\ & \Rightarrow & 2\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}1d\mathrm{\theta }\\ & \Rightarrow & 2\mathrm{I}=\frac{\mathrm{\pi }}{2}-0\\ & \Rightarrow & \mathrm{I}=\frac{\mathrm{\pi }}{4}\end{array}$

Therefore, option (D) is the correct answer.