Question

In your own words, describe how you would integrate \(\int \sin ^{m} x \cos ^{n} x d x\) for each condition. (a) \(m\) is positive and odd. (b) \(n\) is positive and odd. (c) \(m\) and \(n\) are both positive and even.

Short answer

For case (a), use the identity \(\sin^2 x = 1 - \cos^2 x\) and substitute \(u = \cos x\). For case (b), use the identity \(\cos^2 x = 1 - \sin^2 x\) and substitute \(u = \sin x\). For case (c), use power-reducing formulas into basic trigonometric integrals.

Step-by-step solution

Case (a): m is positive and odd

This can be approached by saving one factor of \(\sin x\), then using the identity \(\sin^2 x = 1 - \cos^2 x\). Replace every other factor of \(\sin x\) in the integrand by \(1 - \cos^2 x\) and then make a substitution by letting \(u = \cos x\). The differential \(du\) will be \(-\sin x\, dx\). For the integral, the saved factor of \(\sin x\) allows the \(dx\) to be replaced by \(-du\).

Case (b): n is positive and odd

Similar to Case (a), here one factor of \(\cos x\) should be saved. The identity \(\cos^2 x = 1 - \sin^2 x\) can be employed, replacing every other factor of \(\cos x\) in the integrand by \(1 - \sin^2 x\). Afterwards, the substitution of \(u = \sin x\) can be used, with \(du = \cos x\, dx\). The saved factor of \(\cos x\) then lets the \(dx\) be replaced by \(du\).

Case (c): m and n are both positive and even

In this situation, use the power-reducing formulas \(\cos^2 x = (1 + \cos 2x)/2\) and \(\sin^2 x = (1 - \cos 2x)/2\) to reduce the powers of sine and cosine. The integral will turn into multiples of basic trigonometric integrals.