Finding the integral: $\int_{0}^{\large\frac{\pi}{4}}\frac{\cos(x)\:dx}{a\cos(x)+b \sin(x)}$ What is 
$$\int_{0}^{\large\frac{\pi}{4}}\frac{\cos(x)\:dx}{a\cos(x)+b \sin(x)}?$$
$a,b \in \mathbb{R}$ appropriate fixed numbers. 
 A: Hint. We assume $a>0,b>0$. One may observe that
$$
a\int_{0}^{\large \frac{\pi}{4}}\frac{\cos(x)\:dx}{a\cos(x)+b \sin(x)}+b\int_{0}^{\large \frac{\pi}{4}}\frac{\sin(x)\:dx}{a\cos(x)+b \sin(x)}=\int_0^{\large\frac{\pi}{4}}1\:dx=\frac \pi4
$$ and that
$$
b\int_{0}^{\large \frac{\pi}{4}}\frac{\cos(x)\:dx}{a\cos(x)+b \sin(x)}-a\int_{0}^{\large \frac{\pi}{4}}\frac{\sin(x)\:dx}{a\cos(x)+b \sin(x)}=\int_0^{\large\frac{\pi}{4}}\frac{(a\cos(x)+b \sin(x))'}{a\cos(x)+b \sin(x)}dx.
$$ then solving the system
$$\begin{cases}
a I+bJ=\frac \pi4 \\ 
b I-aJ=\log\left(\frac{a+b}{a \sqrt{2}}\right)
\end{cases}
$$
gives the answer.
A: The substitution $u=\sin(x)$ will help
A: $$\mathcal{I}\left(\text{a},\text{b}\right)=\int_0^{\frac{\pi}{4}}\frac{\cos\left(\text{x}\right)}{\text{a}\cos\left(\text{x}\right)+\text{b}\sin\left(\text{x}\right)}\space\text{d}\text{x}=$$
$$\frac{\text{a}}{\text{a}^2+\text{b}^2}\int_0^{\frac{\pi}{4}}1\space\text{d}\text{x}-\frac{\text{b}}{\text{a}^2+\text{b}^2}\int_0^{\frac{\pi}{4}}\frac{\text{a}\sin\left(\text{x}\right)-\text{b}\cos\left(\text{x}\right)}{\text{a}\cos\left(\text{x}\right)+\text{b}\sin\left(\text{x}\right)}\space\text{d}\text{x}$$
Now, use:
$$\int_0^{\frac{\pi}{4}}1\space\text{d}\text{x}=\frac{\pi}{4}$$
So, we get:
$$\mathcal{I}\left(\text{a},\text{b}\right)=\frac{\pi}{4}\cdot\frac{\text{a}}{\text{a}^2+\text{b}^2}-\frac{\text{b}}{\text{a}^2+\text{b}^2}\int_0^{\frac{\pi}{4}}\frac{\text{a}\sin\left(\text{x}\right)-\text{b}\cos\left(\text{x}\right)}{\text{a}\cos\left(\text{x}\right)+\text{b}\sin\left(\text{x}\right)}\space\text{d}\text{x}$$
Now, substitute $\text{u}=\text{a}\cos\left(\text{x}\right)+\text{b}\sin\left(\text{x}\right)$ and $\text{d}\text{u}=\left(\text{b}\cos\left(\text{x}\right)-\text{a}\sin\left(\text{x}\right)\right)\space\text{d}\text{x}$:
$$\int_0^{\frac{\pi}{4}}\frac{\text{a}\sin\left(\text{x}\right)-\text{b}\cos\left(\text{x}\right)}{\text{a}\cos\left(\text{x}\right)+\text{b}\sin\left(\text{x}\right)}\space\text{d}\text{x}=-\int_\text{a}^{\frac{\text{a}+\text{b}}{\sqrt{2}}}\frac{1}{\text{u}}\space\text{d}\text{u}=-\left(\ln\left|\frac{\text{a}+\text{b}}{\sqrt{2}}\right|-\ln\left|\text{a}\right|\right)$$
So, we get:
$$\mathcal{I}\left(\text{a},\text{b}\right)=\frac{\pi}{4}\cdot\frac{\text{a}}{\text{a}^2+\text{b}^2}+\frac{\text{b}}{\text{a}^2+\text{b}^2}\left(\ln\left|\frac{\text{a}+\text{b}}{\sqrt{2}}\right|-\ln\left|\text{a}\right|\right)$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
&\int_{0}^{\pi/4}{\cos\pars{x} \over a\cos\pars{x} + b\sin\pars{x}}\,\dd x
\\[5mm] = &\
{1 \over a}\int_{0}^{\pi/4}
{\cos\pars{x} \over \cos\pars{x} + \tan\pars{\mu}\sin\pars{x}}\,\dd x
\qquad\qquad\qquad\qquad\pars{~\tan\pars{\mu} \equiv {b \over a}~}
\\[5mm] = &\
{\cos\pars{\mu} \over a}\int_{0}^{\pi/4}
{\cos\pars{x} \over \cos\pars{x - \mu}}\,\dd x =
{\cos\pars{\mu} \over a}\int_{-\mu}^{\pi/4 - \mu}
{\cos\pars{x + \mu} \over \cos\pars{x}}\,\dd x
\\[5mm] = &\
{1 \over a\sec^{2}\pars{\mu}}\,{\pi \over 4} -
{\cos\pars{\mu}\sin\pars{\mu} \over a}\int_{-\mu}^{\pi/4 - \mu}
\tan\pars{x}\,\dd x
\\[5mm] = &
{1 \over a\bracks{\tan^{2}\pars{\mu} + 1}}\,{\pi \over 4} -
{\tan\pars{\mu} \over a\bracks{\tan^{2}\pars{\mu} + 1}}\bracks{-\ln\pars{\cos\pars{{\pi \over 4} - \mu}} + \ln\pars{\cos\pars{-\mu}}}
\\[5mm] = &\
{\pi \over 4}\,{a \over a^{2} + b^{2}} -
{b \over a^{2} + b^{2}}\bracks{%
{1 \over 2}\ln\pars{\tan^{2}\pars{\pi/4 - \mu} + 1 \over
\tan^{2}\pars{\mu} + 1}}
\\[5mm] = &\
{\pi \over 4}\,{a \over a^{2} + b^{2}} -
{b \over 2\pars{a^{2} + b^{2}}}
\ln\pars{\braces{\bracks{1 - b/a}/\bracks{1 + b/a}}^{2} + 1 \over
b^{2}/a^{2} + 1}
\\[5mm] = &\
\bbx{\ds{{\pi \over 4}\,{a \over a^{2} + b^{2}} -
{b \over a^{2} + b^{2}}\ln\pars{\root{2}\verts{a  \over a + b}}}}
\end{align}
