Question

A primitiva de ∫ ( θ - cossecθ cotgθ)dθ é :

a) $\frac{1}{2}$ θ² - cossecθ + c

b) θ² + cossecθ + c

c) $\frac{1}{2}$ θ² + cossecθ + c

d) $\frac{1}{2}$ θ² + secθ + c

e) $\frac{1}{2}$θ² - secθ + c

Answer 1

$\mathrm{\int(\theta-\csc{\theta}\cot{\theta})\ d\theta=\int\theta\ d\theta-\int\csc{\theta}\cot{\theta}\ d\theta=}\\\\ \mathrm{=\dfrac{\theta^2}{2}-\int\dfrac{1}{\sin{\theta}}\dfrac{\cos{\theta}}{\sin{\theta}}\ d\theta=\dfrac{\theta^2}{2}-\int\dfrac{\cos{\theta}}{\sin^2{\theta}}\ d\theta}$

$\mathbf{*\ Resolvendo\ \int\dfrac{\cos{\theta}}{\sin^2{\theta}}\ d\theta:}\\\\ \mathrm{u=\sin{\theta}\ \ \| \ \ \dfrac{du}{dx}=\cos{\theta}\ \to\ du=\cos{\theta}.dx}\\\\ \mathrm{\int\dfrac{1}{\sin^2{\theta}}.\cos{\theta}\ d\theta=\int\dfrac{1}{u^2}\ du=\int u^{-2}\ du=\dfrac{u^{-2+1}}{-2+1}=}\\\\ \mathrm{=\dfrac{u^{-1}}{-1}=-\dfrac{1}{u}=-\dfrac{1}{\sin{\theta}}=-\csc{\theta}}$

$\mathbf{*\ Voltando\ ao\ problema:}\\\\ \mathrm{\int(\theta-\csc{\theta}\cot{\theta})\ d\theta=\dfrac{\theta^2}{2}-(-\csc{\theta})=\boxed{\boxed{\mathbf{\dfrac{\theta^2}{2}+\csc{\theta}}+C}}}$

Resposta: Alternativa C.