Question

$integral \frac{senx + cosx}{ \sqrt[3]{senx - cosx} } dx$

Answer 1

Resposta:

$\boxed{\mathsf{\frac{3}{2} \cdot \sqrt[3]{(\sin x - \cos x)^2} + C}}$

Explicação passo-a-passo:

A resposta sai por substituição simples. Desse modo, tome $\mathbf{\sin x - \cos x = \lambda}$.

Derivando,

$\\ \mathsf{\sin x - \cos x = \lambda} \\\\ \mathsf{[\cos x - (- \sin x)] dx = d\lambda}} \\\\ \mathsf{(\cos x + \sin x) dx = d\lambda}$

Substituindo na integral,

$\\ \displaystyle \mathsf{\int \frac{\sin x + \cos x}{\sqrt[3]{\sin x - \cos x}} dx = \int \frac{d\lambda}{\sqrt[3]{\lambda}}} \\\\\\ \mathsf{= \int \frac{1}{\lambda^{\frac{1}{3}}} d\lambda} \\\\\\ \mathsf{= \int \lambda^{- \frac{1}{3}} d\lambda} \\\\\\ \mathsf{= \frac{3}{2} \cdot \lambda^{- \frac{1}{3} + 1} + C} \\\\\\ \mathsf{= \frac{3}{2} \cdot \lambda^{\frac{2}{3}} + C}$

$\\ \displaystyle \mathsf{= \frac{3}{2} \cdot \sqrt[3]{\lambda^2} + C} \\\\\\ \boxed{\boxed{\mathsf{\int \frac{\sin x + \cos x}{\sqrt[3]{\sin x - \cos x}} dx = \frac{3}{2} \cdot \sqrt[3]{(\sin x - \cos x)^2} + C}}}$