A Special Technique for Integration. Often, two integrands can look similar while one is easy to integrate and the other is very hard. Consider, for example, the integrals : 0/2sinx+cosxsinxdx,0/2sin2x+cos2xsin2xdx,0/2sin10x+cos10xsin10xdx The second one is easy to integrate while the other two are not. Still, you are going to evaluate 0/2sinnx+cosnxsinnxdx where n is a positive integer. To do this you will need a trick and some ingenuity. We will simply provide the trick. Before you work on the entire problem, evaluate the second integral (n=2) without doing anything "hard". Next, use a known technique for integrating rational functions of sinx and cosx to evaluate the first integral. We could use this approach for larger values of n, but imagine how hard it would be. We need to find an easier way. Now here is the trick. Let f be a continuous function over a closed interval [0,a]. Then 0af(x)dx=0af(ax)dx. Prove that this equality is true. Use this result to evaluate 0/2sinnx+cosnxsinnxdx For any value of n. Remember you will need to use some ingenuity here. Make sure your answer agrees with your previous result for n=1 and n=2. Then evaluate the third integral above.