Question

$I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{\cos \theta d\theta }{\left(2-\sin \theta \right)}$

Answer 1

We have,

$I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{\cos \theta d\theta }{\left(2-\sin \theta \right)}$

We know that

${\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$

Therefore,

$I=2{\int }_0^{\frac{\pi }{2}}\frac{\cos \theta d\theta }{\left(2-\sin \theta \right)}$

Let

$t=2-\sin \theta$

$dt=-\cos \theta d\theta$

Therefore,

$I=-2{\int }_2^1\frac{dt}{t}$

$I=-2{\left[\ln \right(t\left)\right]}_2^1$

$I=-2[\ln 1-\ln 2]$

$I=-2(0-\ln 2)$

$I=-2(-\ln 2)$

$I=2\ln 2$

$I=\ln 4$

Hence, this is the answer.