Question

Calcule a integral indefinida:

$\displaystyle\int\frac{\cos x\cos 4x}{\cos 2x+\cos 8x}\,dx$

Answer 1

É dada a seguinte integral indefinida:

$\displaystyle I=\int \dfrac{\cos x\cos 4x }{\cos 2x+\cos 8x}\,dx$

Vamos usar as seguintes relações trigonométricas:

$\cos(p)+\cos(q)=2\cos\left(\dfrac{p+q}{2}\right)\cos\left(\dfrac{p-q}{2}\right)~~~(i)\\\\\ a=\dfrac{p+q}{2},~~b=\dfrac{p-q}{2}\\\\ \Longrightarrow \cos(a)\cos(b)=\dfrac{1}{2}[\cos(a+b)+\cos(a-b)]~~~(ii)$

Aplicando (i) no denominador e (ii) no numerador:

$\displaystyle I=\int \dfrac{\cos x\cos 4x }{\cos 2x+\cos 8x}\,dx\\\\ I=\int \dfrac{\dfrac{1}{2}[\cos(4x+x)+\cos(4x-x)]}{2\cos\left(\dfrac{8x+2x}{2}\right)\cos\left(\dfrac{8x-2x}{2}\right)}\,dx\\\\ I=\dfrac{1}{4}\int \dfrac{\cos(5x)+\cos(3x)}{\cos(5x)\cos(3x)}\,dx\\\\ I=\dfrac{1}{4}\left(\int \dfrac{\cos(5x)}{\cos(5x)\cos(3x)}\,dx+\int \dfrac{\cos(3x)}{\cos(5x)\cos(3x)}\,dx\right)\\\\ I=\dfrac{1}{4}\left(\int \dfrac{1}{\cos(3x)}\,dx+\int \dfrac{1}{\cos(5x)}\,dx\right)$
$\displaystyle I=\dfrac{1}{4}\left(\int \sec(3x)\,dx+\int \sec(5x)}\,dx\right)\\\\ I=\dfrac{1}{4}\left(\dfrac{1}{3}\ln|\sec(3x)+\tan(3x)|+\dfrac{1}{5}\ln|\sec(5x)+\tan(5x)|\right)+C\\\\ \boxed{I=\dfrac{1}{12}\ln|\sec(3x)+\tan(3x)|+\dfrac{1}{20}\ln|\sec(5x)+\tan(5x)|+C}$

Obs: Caso seja necessário explicitar a resolução da integral da secante, basta avisar que farei a edição da resposta.