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Evaluate the integral using substitution

by RoRi

Use the method of substitution to evaluate the following integral:

    \[ \int \frac{(\sin x + \cos x)}{(\sin x - \cos x)^{\frac{1}{3}}} \, dx \]

Let

    \[ u = \sin x - \cos x \quad \implies \quad du = \sin x + \cos x \, dx. \]

Then, we can evaluate the integral,

    \begin{align*} \int \frac{(\sin x + \cos x)}{(\sin x - \cos x)^{\frac{1}{3}}} \, dx &= \int u^{-\frac{1}{3}} \, du \\ &= \frac{3}{2} u^{\frac{2}{3}} + C \\ &= \frac{3}{2} (\sin x - \cos x)^{\frac{2}{3}} + C. \end{align*}

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