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katrin [286]
2 years ago
6

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

Mathematics
1 answer:
masha68 [24]2 years ago
5 0

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 <u>Integration by Substitution</u>

<u />

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}

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