# Trig integration problem

1. Aug 2, 2008

### EWW

Hello everybody,

I've encountered the following problem in Morris Kline's textbook on Calculus (chapter 10, section 5, ex. 2) that I can't seem to understand-

if y' = sin x cos x, then if I set u = sin x, then du/dx = cos x (chain rule), then y = (sin^2 x) / 2. If I set u = cos x, then du/dx= -sin x, so I multiply the RHS of y' by -1/-1 and eventually arrive at y = -(cos^2 x) / 2. Then y = (sin^2 x) / 2 = - (cos^2 x) / 2. What's wrong?

I've been getting other problems in this section, but this one has me stumped . . . . plus I feel that I'm missing something important. Thanks, EW

2. Aug 2, 2008

### slider142

You can't be sure that y = sin^2(x)/2. Rather, you should have y = sin^2(x)/2 + C where C is an undetermined constant, as y' still gives you sin(x)cos(x). Now you have sin^2(x)/2 + C_1 = -cos^2(x)/2 + C_2, which is a true equation. If we carry on, we see that C_2 - C_1 = 1/2. Have you tried graphing these equations?

3. Aug 3, 2008

### schroder

Perhaps the best way to evaluate this function is to use a definite integral, rather than an indefinite with an unknown constant of integration. When you use the definite integral you get (sin^2x1) /2 – (sin^2x2) /2 = - (cos^2x1) /2 – (cos^2x2) /2. You will find that this equality holds for all values of x (in radians).
Note: x1 and x2 are the upper and lower limits of the definite integral.

4. Aug 3, 2008

### EWW

thanks, the equation makes immediate sense to me once I remind myself that sin^2 x + cos^2 x = 1 . . . I will try graphing these because it is still a little strange to me why the constants should be related in some way.