Question

State the strategy of evaluating \(\int {{{\sin }^m}} x{\cos ^n}xdx\)if \(m\) is odd, if \(n\)is odd, and if \(m\)and \(n\)are both even.

Short answer

It is sometimes helpful to use the identity \(\sin x\cos x = \frac{1}{2}\sin 2x\)

Step-by-step solution

Explanation to state the strategy for evaluating \(\int {{{\sin }^m}} x{\cos ^n}xdx\).

Apply the strategy for evaluating the integral function if \(m\), is odd as follows:

If the power if sine is odd \({\rm{(n = 2 k + 1)}}\), save one sine factor and use \({\sin ^2}x = 1 - {\cos ^2}x\)to express the remaining factor in terms of cosine.

\(\begin{array}{l}\int {{{\sin }^{2k + 1}}} x{\cos ^n}xdx = \int {{{\left( {{{\sin }^2}x} \right)}^k}} {\cos ^n}x\sin xdx\\ = \int {{{\left( {1 - {{\cos }^2}x} \right)}^k}} {\cos ^n}x\sin xdx\end{array}\)

Then substitute \(u = \cos x\).

Apply the strategy for evaluating the integral function if \(n\)is odd as follows:

If the power of cosine is odd \({\rm{(n = 2 k + 1)}}\), save one cosine factor and use \({\cos ^2}x = 1 - {\sin ^2}x\)to express the remaining factors in terms of sine.

\(\begin{array}{l}\int {{{\sin }^m}} x{\cos ^{2k + 1}}xdx = \int {{{\sin }^m}} x{\left( {{{\cos }^2}x} \right)^k}\cos xdx\\ = \int {{{\sin }^m}} x{\left( {1 - {{\sin }^2}x} \right)^k}\cos xdx\end{array}\)

Then substitute \(u = \sin x\).

Strategy of evaluating the integral function if \(m\)and \(n\)are both even.

Apply the strategy for evaluating the integral function if \(m\)and \(n\)both are even as follows:

If the powers of both sine and cosine are even, use the half-angle identities.

\(\begin{array}{l}{\sin ^2}x = \frac{1}{2}(1 - \cos 2x)\\{\cos ^2}x = \frac{1}{2}(1 + \cos 2x)\end{array}\)

It is sometimes helpful to use the identity \(\sin x\cos x = \frac{1}{2}\sin 2x\)