Question

Calcule a integral indefinida

     $\displaystyle\int\frac{\cos 10x+\cos 6x}{(1-\cos 10x)(1+\cos 6x)}\,dx$

Dica: Utilize as identidades de prostaférese (transformação de soma em produto) e a identidade para o cosseno do arco duplo.

Answer 1

Olá Lukyo.

Identidade trigonométrica usada:

$\star~~\boxed{\boxed{\mathsf{cos^2t=\dfrac{1}{2}\cdot(1+cos2t)}}}\\\\\\\star~~\boxed{\boxed{\mathsf{sen^2t=\dfrac{1}{2}\cdot(1-cos2t)}}}$

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Organizando a integral.


$\mathsf{\displaystyle\int\dfrac{cos10x +cos6x}{(1-cos10x)(1+cos6x)}~dx}$


Some e subtraia 1 no numerador.

$\mathsf{\displaystyle\int\dfrac{cos10x+cos6x+1-1}{(1-cos10x)(1+cos6x)}~dx}\\\\\\\\\mathsf{\displaystyle\int\dfrac{cos10x+cos6x}{(1-cos10x)(1+cos6x)}~dx=\int\dfrac{(1+cos6x)-(1-cos10x)}{(1-cos10x)(1+cos6x)}~dx}\\\\\\\\\mathsf{\displaystyle\int\dfrac{cos10x+cos6x}{(1-cos10x)(1+cos6x)}~dx=\int\dfrac{(1+cos6x)}{(1-cos10x)(1+cos6x)}~dx-\int\dfrac{(1-cos10x)}{(1-cos10x)(1+cos10x)}~dx}\\\\\\\\\mathsf{\displaystyle\int\dfrac{cos10x+cos6x}{(1-cos10x)(1+cos6x)}~dx=\int\dfrac{1}{(1-cos10x)}~dx-\int\dfrac{1}{(1+cos6x)}~dx}$

Usando a identidade em destaque no ínicio, temos:

$\mathsf{\displaystyle\int\dfrac{cos10x+cos6x}{(1-cos10x)(1+cos6x)}~dx=\int\dfrac{1}{2sen^25x}~dx-\int\dfrac{1}{2cos^23x}~dx}\\\\\\\\\mathsf{\displaystyle\int\dfrac{cos10x+cos6x}{(1-cos10x)(1+cos6x)}~dx=\dfrac{1}{2}\cdot\int cossec^25x~dx-\dfrac{1}{2}\int sec^23x~dx}$

Faça:

$\mathsf{u=5x~~~~~du=5dx}\\\\\\\mathsf{v=3x~~~~~dv=3dx}$

Substituindo:

$\mathsf{\displaystyle\int\dfrac{cos10x+cos6x}{(1-cos10x)(1+cos6x)}~dx=\dfrac{1}{5\cdot2}\cdot\int cossec^2u~du-\dfrac{1}{3\cdot2}\cdot\int sec^2v~dv}\\\\\\\mathsf{\displaystyle\int\dfrac{cos10x+cos6x}{(1-cos10x)(1+cos6x)}~dx=-\dfrac{1}{10}\cdot cotgu-\dfrac{1}{6}\cdot tgv+C}$

Substituindo u e v.

$\mathsf{\displaystyle\int\dfrac{cos10x+cos6x}{(1-cos10x)(1+cos6x)}~dx=-\dfrac{1}{2}\cdot\Big[\dfrac{tg3x}{3}+\dfrac{cotg5x}{5}\Big]+C}$

Dúvidas? Comente.