Question

${\int }_{-\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{3}}\left[\frac{\left(x\sin x\right)}{\left({\cos }^{2}x\right)}\right]dx=$

• A. $\left(\frac{1}{3}\right)\left(4\mathrm{\pi }+1\right)$
• B. $\left[\frac{4\mathrm{\pi }}{3}-2\log \left(\tan \frac{5\mathrm{\pi }}{12}\right)\right]$
• C. $\left[\left(\frac{4\mathrm{\pi }}{3}+\log \tan \left(\frac{5\mathrm{\pi }}{12}\right)\right)\right]$
• D. None of these

Answer 1

The correct option is B

$\left[\frac{4\mathrm{\pi }}{3}-2\log \left(\tan \frac{5\mathrm{\pi }}{12}\right)\right]$

Explanation for the correct option:

Finding the value of the given integral:

Let $I={\int }_{-\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{3}}\left[\frac{\left(x\sin x\right)}{\left({\cos }^{2}x\right)}\right]dx$

$$\begin{gathered} ={\int }_{-\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{3}}\left[x.\frac{\sin x}{\cos x}.\frac{1}{\cos x}\right]dx \\ ={\int }_{-\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{3}}\left[x.secx.\tan x\right]dx \\ =2{\int }_0^{\frac{\mathrm{\pi }}{3}}\left[x.secx.\tan x\right]dx\left\{\mathrm{since}{\int }_{-\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{3}}\left[\frac{\mathrm{xsinx}}{{\cos }^2\mathrm{x}}\right]d\mathrm{x}\mathrm{is}\mathrm{an}\mathrm{even}\mathrm{function}\right\} \end{gathered}$$

By applying $UV$ formula of integration, we get

$$\begin{gathered} =2\left\{{\left[xsecx\right]}_0^{\frac{\mathrm{\pi }}{3}}-\left[{\int }_0^{\frac{\mathrm{\pi }}{3}}1.secxdx\right]\right\} \\ =2\left\{\frac{\mathrm{\pi }}{3}sec\frac{\mathrm{\pi }}{3}-{\left[\log \left(\tan \left(\frac{\mathrm{\pi }}{4}+\frac{x}{2}\right)\right)\right]}_0^{\frac{\mathrm{\pi }}{3}}\right\} \\ =2\left\{\frac{2\mathrm{\pi }}{3}-\log \left(\tan \frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{6}\right)\right\} \\ =\frac{4\mathrm{\pi }}{3}-2\log \left(\tan \frac{5\mathrm{\pi }}{12}\right) \end{gathered}$$

Thus, option (B) is the correct answer.