How do you find the integral of #int sin^n(x)cos^m(x)# if m and n is an integer?

Question

Emilia Aguilar · Accepted Answer

See Explanation

Well that would depend on what #n# and #m# are. There are usually #4# cases to consider: Case 1: If #n# odd. Strip #1# sine out and convert rest to cosines using #sin^2x = 1- cos^2x# , then use the substitution #u = cosx# . Case 2: If #m# is odd, then strip #1# cosine out and convert the rest to sines using #cos^2x=1-sin^2x# and then use the substitution #u=sinx# Case 3: If both #n# and #m# are odd, we can use either method used in Case 1 & 2 Case 4: If both #n# and #m# are even we will need to use double angle and/or half angle formulas to reduce the integral into something we can integrated easier. Check out the following resources for more details and examples: https://tutor.hix.ai Stewart Calculus: Intergrals Involving Trigonometric Functions The following URL leads to Paul's online math notes: tutorial.math.lamar.edu/Classes/CalcII/IntegralsWithTrig.aspx www.math.wisc.edu/~park/Fall2011/integration/Trig%20substitution.pdf is another resource for Trig Substitution.

Evelyn Agard · Answer

undefined To find the integral of $\int \sin^n(x) \cos^m(x) \, dx$ when $m$ and $n$ are integers, we can use the method of integration by parts. We choose $u = \sin^{n-1}(x)$ and $dv = \sin(x) \cos^m(x) \, dx$. Then, $du = (n-1) \sin^{n-2}(x) \cos(x) \, dx$ and $v = \frac{1}{m+1} \cos^{m+1}(x)$.

Applying integration by parts, we have:

$$\int \sin^n(x) \cos^m(x) \, dx = -\frac{\sin^{n-1}(x) \cos^{m+1}(x)}{m+1} + \frac{n-1}{m+1} \int \sin^{n-2}(x) \cos^{m+2}(x) \, dx$$

This process can be repeated iteratively until we reach integrals that can be readily evaluated. Eventually, we obtain an expression involving trigonometric functions and powers of $\sin$ and $\cos$, which may require further simplification or evaluation using trigonometric identities.