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Find indefinite integrals of powers of sin

by RoRi

From the previous two exercises (here and here) we know

    \[ \int \sin^2 x \, dx = \frac{1}{2}x - \frac{1}{4} \sin (2x), \]

and

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

Use these results to establish the following formulas:

  1. \displaystyle{ \int \sin^3 x \, dx = -\frac{3}{4} \cos x + \frac{1}{12} \cos (3x)}.
  2. \displaystyle{ \int \sin^4 x \, dx = \frac{3}{8} x - \frac{1}{4} \sin (2x) + \frac{1}{32} \sin (4x)}.
  3. \displaystyle{ \int \sin^5 x \, dx = -\frac{5}{8} \cos x + \frac{5}{48} \cos (3x) - \frac{1}{80} \cos (5x)}.

    4. ( Note: There is an error in my edition of the book for part (c). It requests we prove \int \sin^5 x \, dx = -\frac{5}{8} x + \frac{5}{48} \cos (3x) - \frac{1}{80} \cos(5x), omitting the \cos x in the first term of the result. This is just an error in the book since the formula as stated is false.)

  1. Using the second formula, and the triple angle identity for cosine (\cos (3x) = 4 \cos^3 x - 3 \cos x which implies \cos^3 x = \frac{1}{4} \cos (3x) + \frac{3}{4} \cos x, derived in this exercise), we have

        \begin{align*} \int \sin^3 x \, dx &= - \frac{\sin^2 x \cos x}{3} + \frac{2}{3} \int \sin x \, dx \\[9pt] &= -\frac{1}{3} \left( \sin^2 x \cos x + 2 \cos x ) \\[9pt] &= -\frac{1}{3} \left( (1 - \cos^2 x) \cos x + 2 \cos x ) \\[9pt] &= -\frac{1}{3} \cos x (3 - \cos^2 x) \\[9pt] &= - \cos x + \frac{1}{3} \cos^3 x \\[9pt] &= - \cos x + \frac{1}{3} \left(\frac{1}{4} \cos (3x) + \frac{3}{4} \cos x \right) \\[9pt] &= -\frac{3}{4} \cos x + \frac{1}{12} \cos (3x). \qquad \blacksquare \end{align*}

  2. Using the second formula above and then the first we have (and also recalling the trig identity \cos (2x) = 1 - 2 \sin^2 x which implies \sin^2 x = \frac{1}{2} - \frac{1}{2} \cos (2x)),

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

  3. Using the second formula above and then part (a) we have

        \begin{align*} \int \sin^5 x \, dx &= -\frac{\sin^4 x \cos x}{5} + \frac{4}{5} \int \sin^3 x \, dx \\[9pt] &= -\frac{1}{5} \left( (1-\cos^2 x)(1-\cos^2 x) \cos x) + \frac{4}{5} \left(-\frac{3}{4} \cos x + \frac{1}{12} \cos (3x) \right) \\[9pt] &= -\frac{1}{5} (\cos x - 2 \cos^3 x + \cos^5 x) + \frac{1}{16} (10 \cos x + 5 \cos (3x) + \cos (5x)) \\[9pt] & \qquad \qquad - \frac{3}{5} \cos x + \frac{1}{15} \cos (3x) \\[9pt] &= -\frac{1}{5} \left( \frac{1}{8} \cos x - \frac{3}{16} \cos (3x) + \frac{1}{16} \cos (5x) \right) - \frac{3}{5}\cos x + \frac{1}{15} \cos (3x) \\[9pt] &= -\frac{5}{8} \cos x + \frac{5}{48} \cos (3x) - \frac{1}{80} \cos (5x). \qquad \blacksquare \end{align*}

2 comments

  1. Eric says:

    So what happens to the arbitrary constant after integrating? It looks like it’s assumed to be 0 in all 3 of these problems.

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