$$\begin{array}{|c|c|} \hline \text{Integral} & \text{Reduction formula} \\ \hline\displaystyle I_{m,n}=\int\sin^max\cos^nax\,\mathrm dx & I_{m,n}=\begin{cases}-\tfrac{\sin^{m-1}ax\cos^{n+1}ax}{a(m+n)}+\tfrac{m-1}{m+n}I_{m-2,n}\\\tfrac{\sin^{m+1}ax\cos^{n-1}ax}{a(m+n)}+\tfrac{n-1}{m+n}I_{m,n-2}\end{cases} \\ \hline \end{array}$$

How to integrate $(\cos x)^n\cdot(\sin x)^m$ with reduction formula. The actual steps please in terms of $(n-2,m)$ and in terms of $(n, m-2)$ like in the table shown. I'm stuck halfway.


  • $\begingroup$ Click enter image description here to view the picture Thanks $\endgroup$ – Natalia Dec 14 '15 at 12:34
  • $\begingroup$ please be more precise about where exactly you are stuck (in the wikipedia article I guess). $\endgroup$ – Surb Dec 14 '15 at 12:44
  • $\begingroup$ See Wallis' integrals. $\endgroup$ – Lucian Dec 14 '15 at 14:29
  • $\begingroup$ @Surb I don't know how to integrate from the beginning. I don't understand how to integrate this equation since there are sin and cos but i understand how to integrate sin^n(x). Pls help I tried so long to understand but I cant. @ Lucian I read it but I don't understand Thanks $\endgroup$ – Natalia Dec 16 '15 at 12:13
  • $\begingroup$ Are you familiar with integration by parts? $\endgroup$ – Simon Jan 2 '16 at 3:17

Consider the example with $m = 3$ and $n= 2$.

We have,

$$I_{m,n} = I_{3,2} = \int \sin^3(ax)\cos^2(ax) \,d x= -\frac{\sin^{3-1}(ax)\cos^{2 + 1}(ax)}{a(3 + 2)} + \frac{3-1}{3 + 2}I_{1,2}.$$

Here I have applied the first reduction formula. I have not reduced the result far enough to be complete. However, I can now continue until the $I_{1,2}$ term disappears. Maybe apply the second reduction next?

Every example would proceed in this way. You must iteratively apply the formula until the integral dissapears or becomes one that you know how to handle.

As an aside for $m \ge 3$ you can also make use of the identity $\sin^2(x) + \cos^2(x) = 1$,

$$\int sin^m(ax) \,d x = \int \sin^{m-2}(ax)(1-\cos^2(ax)) \,d x .$$

In addition, the double-angle substitutions can be useful as well.

These are $$\cos^2(x) = \frac{1}{2}(1 + \cos(2x))$$ and $$ \sin^2(x) = \frac{1}{2}(1 - \cos(2x)).$$


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