Page 214 - Calculus Demystified

CHAPTER 7
EXAMPLE 7.4 Methods of Integration 201
Calculate the integral

2π
sin x cos xdx.
π/2
SOLUTION
We use integration by parts, but we apply the technique to the corresponding
indefinite integral. We let u = sin x and dv = cos xdx. Then

u(x) = sin x du = u (x) dx = cos xdx

v(x) = sin x dv = v (x) dx = cos xdx
So

sin x cos xdx = udv

= uv − vdu

= (sin x) · (sin x) − sin x cos xdx.
At first blush, it appears that we have accomplished nothing. For the new
integral is just the same as the old integral. But in fact we can move the new
integral (on the right) to the left-hand side to obtain

2
2 sin x cos xdx = sin x.
Throwing in the usual constant of integration, we obtain

1
2
sin x cos xdx = sin x + C.
2
Now we complete our work by evaluating the definite integral:
2π 1 1
1
2π
2
2
sin x cos xdx = sin x = sin 2π − sin (π/2) =− .
2
π/2 2 π/2 2 2
We see that there are two ways to treat a definite integral using integration by
parts. One is to carry the limits of integration along with the parts calculation. The
other is to do the parts calculation first (with an indefinite integral) and then plug
in the limits of integration at the end. Either method will lead to the same solution.

You Try It: Calculate the integral

2
e −x cos 2xdx.
0