Question

Evaluate the definite integral from pi/2 to put of cos theta/sqrt 1+ sin theta.​

Answer 1

Answer:

$\textsf{B.}\quad -2(\sqrt{2}-1)$

Step-by-step explanation:

Given integral:

$\displaystyle \int^{\pi}_{\frac{\pi}{2}}\dfrac{\cos \theta}{\sqrt{1+ \sin \theta}}\:\:d\theta$

Solve by using Integration by Substitution

Substitute u for one of the functions of $\theta$ to give a function that's easier to integrate.

$\textsf{Let }u=1+\sin \theta$

Find the derivative of u and rewrite it so that $d \theta$ is on its own:

$\implies \dfrac{du}{d \theta}=\cos \theta$

$\implies d \theta=\dfrac{1}{\cos \theta}\:du$

Use the substitution to change the limits of the integral from $\theta$-values to u-values:

$\textsf{When }\theta=\pi \implies u=1$

$\textsf{When }\theta=\dfrac{\pi}{2} \implies u=2$

Substitute everything into the original integral and solve:

$\begin{aligned}\displaystyle \int^{\pi}_{\frac{\pi}{2}}\dfrac{\cos \theta}{\sqrt{1+ \sin \theta}}\:\:d\theta & =\int^{1}_2}\dfrac{\cos \theta}{\sqrt{u}}\:\cdot \dfrac{1}{\cos \theta}\:\:du\\\\& =\int^{1}_{2}\dfrac{1}{\sqrt{u}} \:\:du \\\\& =\int^{1}_{2} u^{-\frac{1}{2}}\:\:du \\\\& = \left[ 2u^{\frac{1}{2}} \right]^{1}_{2}\\\\& = \left(2(1)^{\frac{1}{2}}\right)-\left(2(2)^{\frac{1}{2}}\right)\\\\& = 2-2\sqrt{2}\\\\& = -2(\sqrt{2}-1)\end{aligned}$