There are two ways to proceed. We can solve this problem by direct calculation: First we compute an antiderivative of , then use it to compute , and finally use this expression to compute .

We have , so , and

.

The second method uses instead the fundamental theorem of calculus directly, and avoids having to compute any antiderivatives: Suppose is a continuous function on the interval and that is an antiderivative of . Recall that this simply means that for .

Suppose is a differentiable function, and is defined by . By the fundamental theorem of calculus, this means that , and therefore

using the chain rule. In the particular case that we are interested in, , , and , so

as before.

Problem 2.Find .

We use the substitution method to simplify the integrand. Although this can be done directly, I will do it in two stages for clarity. First, the most obvious difficulty with the expression at hand is the that appears as the argument in the trigonometric functions. This suggests to first try the substitution

.

Note that , and the integrand already contains a factor of , which is a good sign. To rewrite the integral in terms of , we note that if , then , while if , then , so the integral becomes

.

This expression is still complicated. There are two ways to proceed now. We may use that . This suggests the substitution . Then . When , we have , while when , we have , so the integral becomes

.

The second way to proceed is to use that . This suggests the substitution . Then . When , we have , while when , we have , so the integral becomes

,

as before.

I saw a slightly different presentation of the last approach in some of the solutions: Use the substitution . Then . If , then , while if , then . Then the integral becomes

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