Question

calculo diferencial, alguem sabe a resposta?

Answer 1

Resposta:

Explicação passo-a-passo:

Faça

$u=cos(x)\\\\Assim\\\\du=-sen(x)dx$

Vamos trabalhar com a integral sem os limites de integração e no final voltaremos com eles

$\int{\dfrac{sen(x)}{2cos^{2}x } } \, dx=-\int{\dfrac{-sen(x)}{2cos^{2}x } } \, dx=\\\\\\=-\dfrac{1}{2}.\int{\dfrac{du}{u^{2}}}=-\dfrac{1}{2}.\int{u^{-2}du}=-\dfrac{1}{2}.\dfrac{u^{-2+1}}{(-2+1)}=-\dfrac{1}{2}.\dfrac{u^{-1}}{(-1)}=-\dfrac{1}{2}.\dfrac{(-1)}{u}=\\\\\\=\dfrac{1}{2}.\dfrac{1}{u}=\\\\\\u=cos(x)$

$=\dfrac{1}{2}.[\dfrac{1}{cos(x)}\limits]^\pi _0=\dfrac{1}{2}.[\dfrac{1}{cos(\pi)}-\dfrac{1}{cos(0)}]=\dfrac{1}{2}.[\dfrac{1}{(-1)}-\dfrac{1}{1}]=\\\\\\=\dfrac{1}{2}.[-1}-1]=\dfrac{1}{2}.[-2]=-1$