Encontrar la integral: $\int_{0}^{\large\frac{\pi}{4}}\frac{\cos(x)\:dx}{a\cos(x)+b \sin(x)}$

Question

¿Qué es la $$\int_{0}^{\large\frac{\pi}{4}}\frac{\cos(x)\:dx}{a\cos(x)+b \sin(x)}?$$

$a,b \in \mathbb{R}$ números fijos apropiados.

Answer 1

Sugerencia . Suponemos que $a>0,b>0$ . Se puede observar que $$ a\int_{0}^{\large \frac{\pi}{4}}\frac{\cos(x)\:dx}{a\cos(x)+b \sin(x)}+b\int_{0}^{\large \frac{\pi}{4}}\frac{\sin(x)\:dx}{a\cos(x)+b \sin(x)}=\int_0^{\large\frac{\pi}{4}}1\:dx=\frac \pi4 $$ y que $$ b\int_{0}^{\large \frac{\pi}{4}}\frac{\cos(x)\:dx}{a\cos(x)+b \sin(x)}-a\int_{0}^{\large \frac{\pi}{4}}\frac{\sin(x)\:dx}{a\cos(x)+b \sin(x)}=\int_0^{\large\frac{\pi}{4}}\frac{(a\cos(x)+b \sin(x))'}{a\cos(x)+b \sin(x)}dx. $$ entonces resolviendo el sistema

$$\begin{cases} a I+bJ=\frac \pi4 \\ b I-aJ=\log\left(\frac{a+b}{a \sqrt{2}}\right) \end{cases} $$ da la respuesta.

Answer 2

La sustitución $u=\sin(x)$ ayudará

Answer 3

$$\mathcal{I}\left(\text{a},\text{b}\right)=\int_0^{\frac{\pi}{4}}\frac{\cos\left(\text{x}\right)}{\text{a}\cos\left(\text{x}\right)+\text{b}\sin\left(\text{x}\right)}\space\text{d}\text{x}=$$ $$\frac{\text{a}}{\text{a}^2+\text{b}^2}\int_0^{\frac{\pi}{4}}1\space\text{d}\text{x}-\frac{\text{b}}{\text{a}^2+\text{b}^2}\int_0^{\frac{\pi}{4}}\frac{\text{a}\sin\left(\text{x}\right)-\text{b}\cos\left(\text{x}\right)}{\text{a}\cos\left(\text{x}\right)+\text{b}\sin\left(\text{x}\right)}\space\text{d}\text{x}$$

Ahora, usa:

$$\int_0^{\frac{\pi}{4}}1\space\text{d}\text{x}=\frac{\pi}{4}$$

Entonces, tenemos:

$$\mathcal{I}\left(\text{a},\text{b}\right)=\frac{\pi}{4}\cdot\frac{\text{a}}{\text{a}^2+\text{b}^2}-\frac{\text{b}}{\text{a}^2+\text{b}^2}\int_0^{\frac{\pi}{4}}\frac{\text{a}\sin\left(\text{x}\right)-\text{b}\cos\left(\text{x}\right)}{\text{a}\cos\left(\text{x}\right)+\text{b}\sin\left(\text{x}\right)}\space\text{d}\text{x}$$

Ahora, sustituye $\text{u}=\text{a}\cos\left(\text{x}\right)+\text{b}\sin\left(\text{x}\right)$ y $\text{d}\text{u}=\left(\text{b}\cos\left(\text{x}\right)-\text{a}\sin\left(\text{x}\right)\right)\space\text{d}\text{x}$ :

$$\int_0^{\frac{\pi}{4}}\frac{\text{a}\sin\left(\text{x}\right)-\text{b}\cos\left(\text{x}\right)}{\text{a}\cos\left(\text{x}\right)+\text{b}\sin\left(\text{x}\right)}\space\text{d}\text{x}=-\int_\text{a}^{\frac{\text{a}+\text{b}}{\sqrt{2}}}\frac{1}{\text{u}}\space\text{d}\text{u}=-\left(\ln\left|\frac{\text{a}+\text{b}}{\sqrt{2}}\right|-\ln\left|\text{a}\right|\right)$$

Entonces, tenemos:

$$\mathcal{I}\left(\text{a},\text{b}\right)=\frac{\pi}{4}\cdot\frac{\text{a}}{\text{a}^2+\text{b}^2}+\frac{\text{b}}{\text{a}^2+\text{b}^2}\left(\ln\left|\frac{\text{a}+\text{b}}{\sqrt{2}}\right|-\ln\left|\text{a}\right|\right)$$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}