If , in other words h is defined as the composition of a different function f with yet another function g (and k represents the constant in h), and we know that , then evaluate j(0), if .

Do not use a graphing calculator • Difficulty:

Solution

There are function names all over the place here, but the key function is h, since it is made up of f and g, which in turn make up j, the function we're actually interested in. Luckily, we're also given the derivative of h. So, let's integrate h to determine exactly what f and g are:

The problem tells us that h(x) = f(g(x)) + k, so therefore, f(x) = tan x and g(x) = cos x. That wasn't too hard. Knowing this, we can evaluate j(0) by plugging in 0 everywhere we see an x:

You might be tempted to switch the limits of integration here, since the larger one appears on the as the lower limit. You're definitely allowed to do that, just as long as you multiply the resulting integral by –1. However, you shouldn't. Why? Watch what happens when I do u-substitution again. (To keep from getting confused with my other u-substitution, I am going to use v's for this one.)

Notice that the lower limit of integration (cos 1) is smaller than the upper one (cos 0), so everything's fine! If you're wondering where those new integral limits came from, remember that whenever you do a variable substitution (like we did with v), you need to plug the old limits into that new expression. In this case, I plugged 0 and 1 separately into the equation v = cos t for the t variable. We're almost done:

Remember, cos 1 does not equal 0. There is no simple way to calculate cos 1 without tables or calculators, so this expression cannot be further simplified.