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Prove formulas for the integral of sine and cosine cubed

by RoRi

Prove the following integration formulas:

  1. \displaystyle{\int_0^x \sin^3 t \, dt = \frac{2}{3} - \frac{1}{3} (2 + \sin^2 x) \cos x}.
  2. \displaystyle{\int_0^x \cos^3 t \, dt &= \cos^3 t \, dt = \frac{1}{3} (2 + \cos^2 x) \sin x}.
  1. Proof. First we recall the triple angle identity for sine,

        \[ \sin 3t = 3 \sin t - 4 \sin^3t \quad \implies \quad \sin^3 t = \frac{3}{4} \sin t - \frac{1}{4} sin (3t). \]

    Thus, we have,

        \begin{align*} \int_0^x \sin^3 t \, dt &= \int_0^x \left( \frac{3}{4} \sin t - \frac{1}{4} \sin (3t) \right) \, dt \\ &= \frac{3}{4} (1 - \cos x) - \frac{1}{12} \int_0^{3x} \sin t \, dt \\ &= \frac{3}{4} - \frac{3}{4} \cos x - \frac{1}{12} + \frac{1}{12} \cos 3x \\ &= \frac{2}{3} - \frac{3}{4} \cos x + \frac{1}{12} (\cos x - 4 \sin^2 x \cos x) \\ &= \frac{2}{3} - \frac{2}{3} \cos x - \frac{1}{2} \sin^2 x \cos x \\ &= \frac{2}{3} - \frac{1}{3} (2 + \sin^2 x) \cos x. \qquad \blacksquare \end{align*}

  2. Here we recall the triple angle identity for the cosine

        \[ \cos 3x = 4 \cos^2 x - 3 \cos x. \]

    So we have,

        \begin{align*} \int_0^x \cos^3 t \, dt &= \int_0^x \frac{1}{4} \left( \cos (3t) + 3 \cos t \right) \, dt \\ &= \frac{1}{4} \int_0^x \cos (3t) \, dt + \frac{3}{4} \int_0^x \cos t \, dt \\ &= \frac{1}{12} \left( \left. \sin t \right|_0^{3x} \right) + \frac{3}{4} \left( \left. \sin t \right|_0^x \right) \\ &= \frac{1}{12} ( 3 \sin x - 4 \sin^3 x) + \frac{3}{4} \sin x \\ &= \sin x - \frac{1}{3} \sin^2 x \sin x \\ &= \left( 1 - \frac{1}{3} (1 - \cos^2 x) \right) \sin x \\ &= \left( \frac{2}{3} + \frac{1}{3} \cos^2 x \right) \sin x \\ &= \frac{1}{3} (2 + \cos^2 x) \sin x. \qquad \blacksquare \end{align*}

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