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Prove an identity for integrals of trig functions

by RoRi

For m \in \mathbb{Z}_{>0} prove

    \[ \int_0^{\frac{\pi}{2}} \cos^m x \sin^m x \, dx = 2^{-m} \int_0^{\frac{\pi}{2}} \cos^m x \, dx. \]

Proof. First it will help if we simplify the integral:

    \begin{align*} \int_0^{\frac{\pi}{2}} \cos^m x \sin^m x \,dx &= \int_0^{\frac{\pi}{2}} (\cos x \sin x)^m \, dx \\ &= \int_0^{\frac{\pi}{2}} \left( \frac{1}{2} \sin (2x) \right)^m \, dx \\ &= \frac{1}{2^m} \int_0^{\frac{\pi}{2}} \sin^m (2x) \, dx. \end{align*}

Then we use the method of substitution, letting

    \[ u = 2x, \qquad du = 2 \, dx. \]

So we have,

    \begin{align*} \frac{1}{2^m} \int_0^{\frac{\pi}{2}} \sin^m (2x) \, dx &= \frac{1}{2^{m+1}} \int_{u(0)}^{u(\pi/2)} (\sin u)^m \, du \\[9pt] &= \frac{1}{2^{m+1}} \int_0^{\pi} (\sin u)^m \, du \\[9pt] &= \frac{1}{2^{m+1}} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \sin \left( u + \frac{\pi}{2} \right) \right)^m \, du \\ &= \frac{1}{2^{m+1}} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^m u \, du \\[9pt] &= \frac{1}{2^m} \int_0^{\frac{\pi}{2}} \cos^m x \, dx \\[9pt] &= 2^{-m} \int_0^{\frac{\pi}{2}} \cos^m x \, dx. \qquad \blacksquare \end{align*}

2 comments

  1. Ketan says:

    I was thinking of doing something different:

    Note that 1/2 * integral of cos(x) from 0 to pi/2 = integral of cos(x) * sin(x) from 0 to pi/2 = 1/2 (a)

    Now cos^m(x) / (2^m) – cos^m(x) * sin^m(x) = (cos(x) / 2 – cos(x) * sin(x)) * cos^(m-1) (x) * (1/(2^m-1) + sin(x)/(2^m-2) + sin^2(x)/(2^m-3) + … + sin^(m-1) (x)), where cos^(m-1) (x) and
    (1/(2^m-1) + sin(x)/(2^m-2) + sin^2(x)/(2^m-3) + … + sin^(m-1) (x)) are not negative on the interval from 0 to pi/2.

    If one tries to evaluate the integral of cos^m(x) / (2^m) – cos^m(x) * sin^m(x) then one can use the cauchy-mean value theorem and take out both cos^(m-1) (x) and
    (1/(2^m-1) + sin(x)/(2^m-2) + sin^2(x)/(2^m-3) + … + sin^(m-1) (x)) of the integral, by having them evaluated at some point c, between 0 and pi/2:

    integral of cos^m(x) / (2^m) – cos^m(x) * sin^m(x) from 0 to pi/2
    = cos^(m-1) (c) * (1/(2^m-1) + sin(c)/(2^m-2) + sin^2(c)/(2^m-3) + … + sin^(m-1) (c)) * integral of cos(x) / 2 – cos(x) * sin(x) from 0 to pi/2. From (a), we know that since the integral of cos(x) from 0 to pi/2 = integral of cos(x) * sin(x) from 0 to pi/2 that this is equal to 0, and hence:

    integral of cos^m(x) / (2^m) from 0 to pi/2 = integral of cos^m(x) * sin^m(x) from 0 to pi/2.

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