Question

Calcule a integral indefinida


     $\displaystyle\int\frac{\sin 8x+2\cos 5x+\sin 2x}{\cos 8x+\cos 2x}\,dx$

Answer 1

primeiro vamos simplificar isso ai,
$\frac{ \sin(8x) + 2 \cos(5x) + \ \sin(2x) }{ \cos(8x) + \cos(2x) } = \frac{ 2( \cos(3x). \sin(5x) + \cos(5x) ) }{ 2\cos(5x). \ \cos(3x) } = \frac{ \sin(5x). \cos(3x) + \cos(5x) }{ \cos(5x). \cos(3x) } = \tan(5x) + \sec(3x)$
ai basta calcular a integral de tg(5x)+sec(3x), integral da tg(5x)=(1/5)ln|(sec(5x))|+C e integral de sec(3x)= (1/3)ln|sec(3x)+tg(3x)|+C, então soma as duas e a integral resulta em: (1/5)ln|sec(5x)|+(1/3)ln|sec(3x)+tg(3x)|+C.

Answer 2

Olá Lukyo.


Identidades trigonométricas utilizadas:

$\star~~\boxed{\boxed{\mathsf{sen(\alpha\pm\beta)=sen(\alpha)cos(\beta)\pm sen(\beta)cos(\alpha)}}}\\\\\\\\\star~~\boxed{\boxed{\mathsf{cos(\alpha\pm\beta)=cos(\alpha)cos(\beta)\mp sen(\alpha)sen(\beta)}}}$


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Organizando a integral.


$\mathsf{\displaystyle\int \dfrac{sen(8x)+2cos(5x)+sen(2x)}{cos(8x)+cos(2x)}~dx}$


Pela identidade do seno destacada no ínicio, temos que:


$\mathsf{sen(3x+5x)=sen(5x)cos(3x)+sen(3x)cos(5x)~ (i)}\\\\\mathsf{sen(5x-3x)=sen(5x)cos(3x)-sen(3x)cos(5x)~(ii)}$

Somando (i) e (ii), temos:

$\mathsf{sen(8x)+sen(2x)=2sen(5x)cos(3x)}$


Agora pela identidade do cosseno:


$\mathsf{cos(5x+3x)=cos(5x)cos(3x)-sen(5x)sen(3x)~(i)}\\\\\\\mathsf{cos(5x-3x)=cos(5x)cos(3x)+sen(5x)sen(3x)~(ii)}$


Somando (i) e (ii), temos:

$\mathsf{cos(8x)+cos(2x)=2cos(5x)cos(3x)}$


Substituindo na nossa integral.


$\mathsf{\displaystyle\int\dfrac{sen(8x)+2cos(5x)+sen(2x)}{cos(8x)+cos(2x)}~dx=\int\dfrac{2sen(5x)cos(3x)+2cos(5x)}{2cos(5x)cos(3x)}~dx}\\\\\\\mathsf{\displaystyle\int\dfrac{sen(8x)+2cos(5x)+sen(2x)}{cos(8x)+cos(2x)}~dx=\int\Big[\dfrac{2sen(5x)cos(3x)}{2cos(5x)cos(3x)}+\dfrac{2cos(5x)}{2cos(5x)cos(3x)}\Big]~dx}\\\\\\\mathsf{\displaystyle\int\dfrac{sen(8x)+2cos(5x)+sen(2x)}{cos(8x)+cos(2x)}~dx=\int\Big[\dfrac{sen(5x)}{cos(5x)}+\dfrac{1}{cos(3x)}\Big]~dx}$

$\mathsf{\displaystyle\int\dfrac{sen(8x)+2cos(5x)+sen(2x)}{cos(8x)+cos(2x)}~dx=\int [tg(5x)+sec(3x)]~dx}$

Faça:

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


$\mathsf{\displaystyle\int\dfrac{sen(8x)+2cos(5x)+sen(2x)}{cos(8x)+cos(2x)}~dx=\dfrac{1}{5}\cdot\int tg(u)~du+\dfrac{1}{3}\cdot\int sec(v)~dv}\\\\\\\mathsf{\displaysytle\int\dfrac{sen(8x)+2cos(5x)+sen(2x)}{cos(8x)+cos(2x)}~dx=-\dfrac{ln|cos(u)|}{5}+\dfrac{ln|tan(v)+sec(v)|}{3}+C}$


Substituindo u e v.


$\mathsf{\displaystyle\int\dfrac{sen(8x)+2cos(5x)+sen(2x)}{cos(8x)+cos(2x)}~dx=\dfrac{ln|tg(3x)+sec(3x)|}{3}-\dfrac{ln|cos(5x)|}{5}+C}$


Dúvidas? Comente.