Problem source

Let
\[
I= \int_0^a \frac {\cos x}{\sin x + \cos x} \; \d x \,
\quad \mbox{ and  } \quad
J= \int_0^a \frac {\sin  x}{\sin x + \cos x} \; \d x \;,
\]
where $0\le a < \frac{3}{4}\pi\,$.
By considering $I+J$ and $I-J$,  show that 
$
2I= a + \ln (\sin a +\cos a)\;.
$
Find also:
\begin{questionparts}
\item 
$\displaystyle \int_0^{\frac{1}{2}\pi} \frac {\cos x}{p\sin x + q\cos x} \; \d x \,$, where $p$ and $q$ 
are positive numbers;
%\item [(ii)] 
%$\displaystyle \int_0^{\frac{1}{2}\pi/2} \frac {\cos x}{\sin (x+k)} \; \d x \,$, where $0 < k < \pi/2\,$;
\item 
$\displaystyle \int_0^{\frac{1}{2}\pi} \frac {\cos x+4}{3\sin x + 4\cos x+ 25} \; \d x \,$.
\end{questionparts}

Solution source

\begin{align*}
&& I + J &= \int_0^a \frac{\sin x + \cos x}{\sin x + \cos x } \d x = a \\
&& I - J &= \int_0^a \frac{\cos x - \sin x}{\sin x + \cos x} \d x \\
&&&= \left [\ln ( \sin x + \cos x) \right]_0^a = \ln (\sin a + \cos a) - \ln 1 = \ln(\sin a + \cos a) \\
\\
\Rightarrow && 2I &= a + \ln(\sin a + \cos a)
\end{align*}

\begin{questionparts}
\item Let $\displaystyle I = \int_0^{\frac12 \pi} \frac{\cos x}{p \sin x + q \cos x} \d x, J = \int_0^{\frac12 \pi} \frac{\sin x}{p \sin x + q \cos x} \d x$ so

\begin{align*}
&& qI + pJ &= \frac{\pi}{2} \\
&& pI - qJ &= \int_0^{\frac12 \pi} \frac{p \cos x - q \sin x}{p \sin x + q \cos x } \d x \\
&&&= \left [\ln (p \sin x + q \cos x) \right]_0^{\pi/2} \\
&&&= \ln(p) - \ln(q) = \ln \frac{p}{q}
\end{align*}

\item $\,$
\begin{align*}
&& \int_0^{\frac{\pi}{2}} \frac{\cos x}{\sin(x + k)} \d x &= \int_0^{\frac{\pi}{2}} \frac{\cos x}{\sin(x) \cos(k) + \cos(x) \sin (k)} \d x \\
&&&=  \ln \tan k
\end{align*}

\item Let $\displaystyle I =  \int_0^{\pi/2} \frac{\cos x + 4}{3 \sin x + 4 \cos x + 25} \d x, J = \int_0^{\pi/2} \frac{\sin x + 3}{3 \sin x + 4 \cos x + 25} \d x$, so

\begin{align*}
&& 4I + 3J &= \int_0^{\pi/2} \frac{3 \sin x + 4 \cos x + 25}{3 \sin x + 4 \cos x + 25} \d x \\
&&&= \frac{\pi}{2} \\
&& 3I - 4J &= \int_0^{\pi/2} \frac{3\cos x - 4 \sin x}{3 \sin x + 4 \cos x + 25} \d x \\
&&&= \left [\ln(3 \sin x + 4 \cos x + 25) \right]_0^{\pi/2} \\
&&&= \ln (28) - \ln (29) = \ln \frac{28}{29} \\
\Rightarrow && 25I &= 2\pi + 3 \ln \frac{28}{29} \\
\Rightarrow && I &= \frac{2}{25} \pi + \frac{3}{25} \ln \frac{28}{29}
\end{align*}
\end{questionparts}