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Find formulas for the integral of powers of cosine

From the previous exercise we know

    \[ \int \cos^n x \, dx = \frac{\cos^{n-1} x \sin x}{n} + \frac{n-1}{n} \int \cos^{n-2} x \, dx. \]

Use this to establish the following:

  1. \displaystyle { \cos^2 x\, dx = \frac{1}{2} x + \frac{1}{4} \sin (2x)}.
  2. \displaystyle { \cos^3 x\, dx = \frac{3}{4} x + \frac{1}{12} \sin (3x)}.
  3. \displaystyle { \cos^4 x\, dx = \frac{3}{8} x + \frac{1}{4} \sin (2x) + \frac{1}{32} \sin (4x)}.

  1. Proof. Applying the solution from the previous exercise and the trig identity \sin (2x) = 2 \sin x \cos x we have,

        \begin{align*}  \int \cos^2 x \, dx &= \frac{\cos x \sin x}{2} + \frac{1}{2} \int dx \\[9pt]  &= \frac{1}{2} x + \frac{1}{4} \sin (2x). \qquad \blacksquare \end{align*}

  2. Proof. Again we apply the solution from the previous exercise, and this time we use the pythagorean identity \cos^2x = 1 - \sin^2 x to obtain

        \begin{align*}  \int \cos^3 x \, dx &= \frac{\cos^2 x \sin x}{3} + \frac{2}{3} \int \cos x \, dx \\[9pt]   &= \frac{2}{3} \sin x + \frac{\sin x - \sin^3 x}{3} \\[9pt]  &= \sin x - \frac{1}{3} \sin^3 x. \qquad \blacksquare \end{align*}

  3. Proof. We apply the solution from the previous problem, the solution of part (a), and trig identities to derive the following:

        \begin{align*}  \int \cos^4 x \, dx &= \frac{\cos^3 x \sin x }{4} + \frac{3}{4} \int \cos^2 x \, dx \\[9pt]  &= \frac{1}{4} \left( \frac{1}{2} \sin (2x) \cos^2 x \right) + \frac{3}{4} \left( \frac{1}{2} x + \frac{1}{4} \sin (2x) \right) \\[9pt]  &= \frac{3}{8}x + \frac{3}{16} \sin (2x) + \frac{1}{8} (\sin (2x) - \sin (2x) \sin^2 x) \\[9pt]  &= \frac{3}{8}x + \frac{5}{16} \sin (2x) - \frac{1}{16} (\sin (2x) - \sin (2x) \cos (2x)) \\[9pt]  &= \frac{3}{8}x + \frac{1}{4} \sin (2x) + \frac{1}{32} \sin (4x). \qquad \blacksquare \end{align*}

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