It's just missing a cosine

Call the first integral (if taken without bounds) $I$ . There's a nice way to do this if you consider a "counterpart" integral, $J=\int _{}^{}{ \frac { \cos{x} }{ \sin { x } +\cos { x } } } dx$ Then we have $I+J=x$ . Observe that $I-J$ is a far easier integral:

$I-J=\int _{ }^{ }{ \frac { \sin { x } -\cos { x } }{ \sin { x } +\cos { x } } } dx$

then substitute $u=\sin{x}+\cos{x} \Rightarrow dx= \frac{1}{\cos{x}-\sin{x}} du$ . This can cancel! We get

$I-J= - \int_{}^{}{\frac{1}{u}} du = -\ln{|\sin{x}+\cos{x}|}$

Thus, $\frac{(I+J)+(I-J)}{2}=I=\frac{x-\ln{|\sin{x}+\cos{x}|}}{2}$ . Plugging in the bounds, in both cases the logarithm becomes $0$ and we are left with $\frac{\pi}{4} \approx 0.7854$ .

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While this gives the indefinite integral, there is a faster way to solve the indefinite integral posed. (Credit to Pi Han Goh):

Take $J$ and substitute $u=\frac{\pi}{2}-x$ . Then for the given bounds, $J=I$ and thus by adding $I$ and $J$ we get

$2I = \int _{0}^{\frac{\pi}{2}}{1} dx \Rightarrow I=\frac{\pi}{4}$