Question

Evaluate :${\int }_0^{\pi }[cotx]dx$

• A. $\pi /2$
• B. $1$
• C. $-1$
• D. $-\left(\pi /2\right)$

Answer 1

The correct option is D

$-\left(\pi /2\right)$

Step 1:Consider the given equation as:

$I={\int }_0^{\pi }\left[cotx\right]dx......\left(1\right)$

we know the property of definite integral.

${\int }_0^{a}xdx={\int }_0^a\left(a-x\right)dx$

similarly,

$$\begin{gathered} I={\int }_0^{\pi }\left[cot\left(\pi -x\right)\right]dx \\ I={\int }_0^{\pi }\left[-cot\left(x\right)\right]dx.......\left(2\right) \end{gathered}$$

Since, $cot\left(\pi -x\right)=-cotx$

Step 2: Adding the Equation $\left(1\right)\mathrm{and}\left(2\right)$ to get the value of $I$.

$$\begin{gathered} 2I={\int }_0^{\pi }\left[cotx\right]dx+{\int }_0^{\pi }\left[-cotx\right]dx \\ 2I={\int }_0^{\pi }\left(-1\right)dx \end{gathered}$$

We know that $$\begin{gathered} \left[x\right]+\left[-x\right]=-1ifx\nexists z \\ \left[x\right]+\left[-x\right]=0ifx\in z \end{gathered}$$

$$\begin{gathered} 2I={\left[-x\right]}_0^{\pi } \\ 2I=-\pi \\ I=-\frac{\pi }{2} \end{gathered}$$

Hence, the correct answer is Option D.