Question

If ${\theta }_1$ and ${\theta }_2$ be respectively the smallest and the largest values of $\theta$ in $(0,2\pi )-\left\{\pi \right\}$which satisfy the equation $2co{t}^2\theta -\frac{5}{\sin \theta }+4=0$

then ${\int }_{{\theta }_1}^{{\theta }_2}{\cos }^{2}3\theta d\theta$ is equal to:

• A. $\frac{2\mathrm{\pi }}{3}$
• B. $\frac{\mathrm{\pi }}{3}$
• C. $\left(\frac{\mathrm{\pi }}{3}\right)+\left(\frac{1}{6}\right)$
• D. $\frac{\mathrm{\pi }}{9}$

Answer 1

The correct option is B

$\frac{\mathrm{\pi }}{3}$

Explanation for the correct option.

Step 1. Find the solutions of the equation.

Let's consider the given function

$$\begin{gathered} 2{\cot }^2\theta -\frac{5}{\sin \theta }+4=0 \\ \Rightarrow 2\left({\csc }^2\theta -1\right)-5\csc \theta +4=0\left[\because {\cot }^2\theta ={\csc }^2\theta -1;\frac{1}{\mathrm{sin\theta }}=\mathrm{csc\theta }\right] \\ \Rightarrow 2{\csc }^2\theta -4\csc \theta -\csc \theta +2=0 \\ \Rightarrow 2\csc \theta (\csc \theta -2)-\left(\csc \theta -2\right)=0 \\ \Rightarrow \left(\csc \theta -2\right)\left(2\csc \theta -1\right)=0 \end{gathered}$$

So either $\csc \theta =2$ or $\csc \theta =\frac{1}{2}$. But $\csc \theta =\frac{1}{2}$ is not possible.

So for $\theta \in \left(0,2\mathrm{\pi }\right)$ the values of $\theta$ for which $\csc \theta =2$ are ${\theta }_1=\frac{\mathrm{\pi }}{6}$ and ${\theta }_2=\frac{5\mathrm{\pi }}{6}$.

Step 2. Find the value of the integral.

The integral limit has been found so it can be evaluated as:

$\begin{array}{rcl}{\int }_{{\theta }_1}^{{\theta }_2}{\cos }^{2}3\theta d\theta & =& {\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{5\mathrm{\pi }}{6}}{\cos }^{2}3\theta d\theta \\ & =& {\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{5\mathrm{\pi }}{6}}\left(\frac{1+\cos 6\theta }{2}\right)d\theta \\ & =& \frac{1}{2}{\left[\theta +\frac{\sin 6\theta }{6}\right]}_{\frac{\mathrm{\pi }}{6}}^{\frac{5\mathrm{\pi }}{6}}\\ & =& \frac{1}{2}\left[\frac{5\mathrm{\pi }}{6}-\frac{\mathrm{\pi }}{6}+\frac{1}{6}\left(\sin \left(6\times \frac{5\mathrm{\pi }}{6}\right)-\sin \left(6\times \frac{\mathrm{\pi }}{6}\right)\right)\right]\\ & =& \frac{1}{2}\left[\frac{4\mathrm{\pi }}{6}+\frac{1}{6}\left(0-0\right)\right]\\ & =& \frac{\mathrm{\pi }}{3}\end{array}$

Hence, the correct option is B.