Question

The value of ${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+\cot \left(x\right)}dx$is

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

Answer 1

The correct option is D

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

Explanation for correct option

Given: ${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+\cot \left(x\right)}dx$

Let $$\begin{gathered} I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+{\displaystyle \frac{\cos \left(x\right)}{\sin \left(x\right)}}}dx \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin \left(x\right)}{\sin \left(x\right)+{\displaystyle \cos \left(x\right)}}dx--------\left(1\right) \end{gathered}$$

using the property $x\to a+b-x$ ${\int }_b^{a}x·dx={\int }_b^a(a+b-x)·dx$

$$\begin{gathered} I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin \left(\frac{\mathrm{\pi }}{2}+0-x\right)}{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}+0-x\right)+{\displaystyle \cos \left(\frac{\mathrm{\pi }}{2}+0-x\right)}}dx \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)+\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}dx \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\cos \left(x\right)}{\cos \left({\displaystyle x}\right)+\sin \left(x\right)}dx--------\left(2\right) \end{gathered}$$

adding $\left(1\right)$ and $\left(2\right)$

$$\begin{gathered} \Rightarrow I\mathit{+}I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\cos \left(x\right)}{\cos \left({\displaystyle x}\right)+\sin \left(x\right)}dx+{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin \left(x\right)}{\sin \left({\displaystyle x}\right)+\cos \left(x\right)}dx \\ \Rightarrow 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\cos \left(x\right)+\sin \left(x\right)}{\cos \left({\displaystyle x}\right)+\sin \left(x\right)}dx \\ \Rightarrow 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}1dx\left[\int c·dx=cx\right] \\ \Rightarrow 2I={\left|x\right|}_0^{\frac{\mathrm{\pi }}{2}} \\ \Rightarrow 2I=\left|\frac{\mathrm{\pi }}{2}-0\right| \\ \Rightarrow 2I=\frac{\mathrm{\pi }}{2} \\ \Rightarrow I=\frac{\mathrm{\pi }}{4} \end{gathered}$$

Hence, option D is correct.