Question

A alternativa que corresponde a solução de: ∫sen(4x)sen(3x)ds definida em -pi/2 a pi/2 É:

Answer 1

Explicação passo-a-passo:

Integral definida

Dada a integral :

$\red{ \displaystyle\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \sf{ \sin(4x)*\sin(3x) dx } }$

Como: $\sf{\green{ \sin(\alpha) *\cos(b)~=~ \dfrac{1}{2}*\left( \sin(\alpha - b) + \cos( \alpha + b) \right) } }$

Então :

$\sf{I~=~} \displaystyle\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \sf{ \dfrac{1}{2}\left( \cos(4x - 3x) + \cos( 4x + 3x) \right)dx}$

Vamos primeiro achar a indefinida :

$\sf{I~=~} \left( \sf{\dfrac{1}{2}}\displaystyle\int \sf{\left( \cos(x) + \cos(7x) \right)dx }\right)\Bigg|^{\frac{\pi}{2}}_{-\frac{\pi}{2}}$

$\sf{I~=~} \left( \sf{\dfrac{1}{2}}\left( \displaystyle\int \sf{\cos(x)dx} + \int\sf{\cos(7x)dx} \right) \right)\Bigg|^{\frac{\pi}{2}}_{-\frac{\pi}{2}}$

$\sf{I~=~ } \left( \sf{\dfrac{1}{2}}\left( \displaystyle\int\sf{cos(x)dx} \sf{+\dfrac{1}{7}} \int \sf{ 7\cos(7x)dx} \right) \right)\Bigg|^{\frac{\pi}{2}}_{-\frac{\pi}{2}}$

$\sf{I~=~ \left( \dfrac{1}{2}*\left( \sin(x) + \dfrac{\sin(7x)}{7} \right) \right)\Bigg|^{\frac{\pi}{2}}_{-\frac{\pi}{2}} }$

$\sf{I~=~\left( \dfrac{\sin(x)}{2} + \dfrac{\sin(7x)}{14}\right) \Bigg|^{\frac{\pi}{2}}_{-\frac{\pi}{2}} }$

$\sf{ I~=~ \dfrac{ \sin\left( \frac{\pi}{2} \right) }{2} + \dfrac{ \sin\left( \frac{7\pi}{2} \right) }{14} - \left( \dfrac{ \sin\left( -\frac{\pi}{2}\right) }{2} + \dfrac{ \sin\left(-\frac{7\pi}{2} \right) }{ 14 } \right) }$

$\sf{ I~=~ \dfrac{ 1 }{2} + \dfrac{1}{14} - \left( -\dfrac{1}{2} - \dfrac{1}{14} \right) }$

$\sf{ I~=~ \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{14} + \dfrac{1}{14} }$

$\iff \sf{ I~=~ \dfrac{7}{7} + \dfrac{1}{7} }$

$\pink{ \iff \boxed{ \sf{ I~=~ \dfrac{8}{7} } } \sf{ \longleftarrow Resposta } }$

Alternativa É)

Espero ter ajudado bastante!)