Answer:
[tex]\textsf{B.}\quad -2(\sqrt{2}-1)[/tex]
Step-by-step explanation:
Given integral:
[tex]\displaystyle \int^{\pi}_{\frac{\pi}{2}}\dfrac{\cos \theta}{\sqrt{1+ \sin \theta}}\:\:d\theta[/tex]
Solve by using Integration by Substitution
Substitute u for one of the functions of [tex]\theta[/tex] to give a function that's easier to integrate.
[tex]\textsf{Let }u=1+\sin \theta[/tex]
Find the derivative of u and rewrite it so that [tex]d \theta[/tex] is on its own:
[tex]\implies \dfrac{du}{d \theta}=\cos \theta[/tex]
[tex]\implies d \theta=\dfrac{1}{\cos \theta}\:du[/tex]
Use the substitution to change the limits of the integral from [tex]\theta[/tex]-values to u-values:
[tex]\textsf{When }\theta=\pi \implies u=1[/tex]
[tex]\textsf{When }\theta=\dfrac{\pi}{2} \implies u=2[/tex]
Substitute everything into the original integral and solve:
[tex]\begin{aligned}\displaystyle \int^{\pi}_{\frac{\pi}{2}}\dfrac{\cos \theta}{\sqrt{1+ \sin \theta}}\:\:d\theta & =\int^{1}_2}\dfrac{\cos \theta}{\sqrt{u}}\:\cdot \dfrac{1}{\cos \theta}\:\:du\\\\& =\int^{1}_{2}\dfrac{1}{\sqrt{u}} \:\:du \\\\& =\int^{1}_{2} u^{-\frac{1}{2}}\:\:du \\\\& = \left[ 2u^{\frac{1}{2}} \right]^{1}_{2}\\\\& = \left(2(1)^{\frac{1}{2}}\right)-\left(2(2)^{\frac{1}{2}}\right)\\\\& = 2-2\sqrt{2}\\\\& = -2(\sqrt{2}-1)\end{aligned}[/tex]
