Answer
$\ln(3+2\sqrt{2})$
Work Step by Step
$A=\displaystyle \int_{a}^{b}f(x)dx$
$A=\displaystyle \int_{\pi/4}^{3\pi/4}\frac{\sin x}{1+\cos x}dx$
... Find an antiderivative (evaluate the indefinite integral)
$\left[\begin{array}{l}
u=1+\cos x\\
du=-\sin x
\end{array}\right]$
$\displaystyle \int_{ }^{ }\frac{\sin x}{1+\cos x}dx=\int_{ }^{ }\frac{-1}{u}dx$
$=-\ln|u|+C=-\ln|1+\cos x|+C$
$A=[-\ln|1+\cos x|]_{\pi/4}^{3\pi/4}=$
$=-\displaystyle \ln|1+\cos\frac{3\pi}{4}|+\ln|1+\cos\frac{\pi}{4}|$
$=-\displaystyle \ln|1-\frac{\sqrt{2}}{2}|+\ln|1+\frac{\sqrt{2}}{2}|$
... quotient rule ...
$=\ln\left(\dfrac{\frac{2+\sqrt{2}}{2}}{\frac{2-\sqrt{2}}{2}}\right)=$
$=\displaystyle \ln\frac{2+\sqrt{2}}{2-\sqrt{2}}$
$\displaystyle \frac{2+\sqrt{2}}{2-\sqrt{2}}\cdot\frac{2+\sqrt{2}}{2+\sqrt{2}} =\displaystyle \frac{4+2\cdot 2\cdot\sqrt{2}+2}{4-2}=\frac{6+4\sqrt{2}}{2}=3+2\sqrt{2}$
$A=\ln(3+2\sqrt{2})$
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