Question

${\int }_{-\pi /2}^{\pi /2}ln\left[2\left(\frac{4-\sin \theta }{4+\sin \theta }\right)\right]d\theta$.

Answer 1

Given : ${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\ln \left|2\left(\frac{4-\sin \theta }{4+\sin \theta }\right)\right|d\theta .........\left(1\right)$

$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\ln \left[2\left|\frac{4-\sin \theta }{4+\sin \theta }\right|\right]d\theta (replace\theta =\pi /2-\pi /2-\theta )I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\ln \left[2\left|\frac{4+\sin \theta }{4-\sin \theta }\right|\right]d\theta ...........\left(2\right)$

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

$2I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}[\ln \left[2\left|\frac{4-\sin \theta }{4+\sin \theta }\right|\right]+\ln \left[2\left|\frac{4+\sin \theta }{4-\sin \theta }\right|\right]]d\theta 2I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\ln |\frac{4-\sin \theta }{4+\sin \theta }\times \frac{4+\sin \theta }{4-\sin \theta }|d\theta 2I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\ln 4d\theta =\ln 4{\left[\theta \right]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}2I=\ln 4(\frac{\pi }{2}+\frac{\pi }{2})=\left(\ln 4\pi \right)I=\left(\ln 2\right)\pi$

Hence the correct answer is $\left(\ln 2\right)\pi$