Question

resolver a integral pelo método da substituição.​​

Answer 1

Resposta:

$\int\frac{\sin\theta}{(5-\cos\theta)^3}\;d\theta=-\frac{1}{2(5-\cos\theta)^2}+C$

Explicação passo-a-passo:

Considerando $5-\cos\theta=u$, temos que:

$\frac{du}{d\theta}=\frac{d}{d\theta}(5-\cos\theta)$

$\frac{du}{d\theta}=\frac{d}{d\theta}(5)+\frac{d}{d\theta}(-\cos\theta)$

$\frac{du}{d\theta}=0-\frac{d}{d\theta}(\cos\theta)$

$\frac{du}{d\theta}=-(-\sin\theta)$

$\frac{du}{d\theta}=\sin\theta$

$du=\sin\theta\;d\theta$

Dessa forma, temos que:

$\int\frac{\sin\theta}{(5-\cos\theta)^3}\;d\theta=\int\frac{1}{u^3}*\sin\theta\;d\theta$

$\int\frac{\sin\theta}{(5-\cos\theta)^3}\;d\theta=\int\frac{1}{u^3}\;du$

$\int\frac{\sin\theta}{(5-\cos\theta)^3}\;d\theta=\int u^{-3}\;du$

$\int\frac{\sin\theta}{(5-\cos\theta)^3}\;d\theta=\frac{u^{-3+1}}{-3+1}+C$

$\int\frac{\sin\theta}{(5-\cos\theta)^3}\;d\theta=\frac{u^{-2}}{-2}+C$

$\int\frac{\sin\theta}{(5-\cos\theta)^3}\;d\theta=-\frac{1}{2u^2}+C$

$\int\frac{\sin\theta}{(5-\cos\theta)^3}\;d\theta=-\frac{1}{2(5-\cos\theta)^2}+C$