Section 7.2, Integration by Parts

Homework: 7.2 #1–57 odd

2 So far, we have been able to integrate some products, such as R xex dx. They have been able to be solved by a u-substitution. However, what happens if we can’t solve it by a u-substitution? For example, consider R xex dx. For this, we will need a technique called integration by parts. 0 0 From the product rule for derivatives, we know that Dx[u(x)v(x)] = u (x)v(x) + u(x)v (x). Rear- ranging this and integrating, we see that:

0 0 u(x)v (x) = Dx[u(x)v(x)] − u (x)v(x) Z Z Z 0 0 u(x)v (x) dx = Dx[u(x)v(x)] dx − u (x)v(x) dx

This gives us the formula needed for integration by parts: Z Z u(x)v0(x) dx = u(x)v(x) − u0(x)v(x) dx, or, we can write it as Z Z u dv = uv − v du

In this section, we will practice choosing u and v properly. Examples Perform each of the following integrations: 1. R xex dx Let u = x, and dv = ex dx. Then, du = dx and v = ex. Using our formula, we get that Z Z xex dx = xex − ex dx = xex − ex + C

R ln√x 2. x dx Since we do not know how to integrate ln x, let u = ln x and dv = x−1/2. Then du = 1/x and v = 2x1/2, so Z ln x Z √ dx = 2x1/2 ln x − 2x−1/2 dx x = 2x1/2 ln x − 4x1/2 + C

3. R arctan x dx We aren’t given a (obvious) product here, but we don’t have an integration formula for arctan x. So, we can think of the integral as R 1 · arctan x dx, and let u = arctan x and dv = 1 dx. Then, dx du = 1+x2 and v = x. Using the formula, Z Z x arctan x dx = x arctan x − dx 1 + x2 1 = x arctan x − ln(1 + x2) + C 2 A similar technique can be used to integrate other inverse trigonometric functions, as well as the natural logarithm function.

1 4. R x2 sin(2x) dx 2 cos(2x) Let u = x and dv = sin(2x) dx. Then, du = 2x dx and v = − 2 . Using our formula, Z x2 Z x2 sin(2x) dx = − cos(2x) + x cos(2x) dx 2

Since we do not know the integral of x cos(2x), we will repeat integration by parts with u = x sin(2x) and dv = cos(2x) dx. Then, du = dx and v = 2 . We then get Z x Z sin(2x) x cos(2x) dx = sin(2x) − dx 2 2 x cos(2x) = sin(2x) + + C 2 4 As a result, our final answer is

Z x2 x cos(2x) x2 sin(2x) dx = − cos(2x) + sin(2x) + + C 2 2 4

5. R ex cos x dx Let u = ex and dv = cos x dx. Then, du = ex dx and v = sin x, so Z Z ex cos x dx = ex sin x − ex sin x dx

We will repeat integration by parts with u = ex and dv = sin x dx, so du = ex dx and v = − cos x. From our formula, we see: Z Z ex cos x dx = ex sin x − ex sin x dx Z = ex sin x + ex cos x − ex cos x dx. Solving for the integral, we get Z 2 ex cos x dx = ex sin x + ex cos x Z 1 1 ex cos x dx = ex sin x + ex cos x + C 2 2

6. Derive a reduction formula for R cosn x dx when n ≥ 2. Let u = cosn−1 x and dv = cos x dx. Then, du = −(n − 1) cosn−2 x sin x dx and v = sin x. By our formula, Z Z cosn x dx = sin x cosn−1 x + (n − 1) cosn−2 x sin2 x dx Z = sin x cosn−1 x + (n − 1) cosn−2 x(1 − cos2 x) dx Z Z = sin x cosn−1 x + (n − 1) cosn−2 x dx − (n − 1) cosn x dx.

Solving for original integral, we get that Z Z n cosn x dx = sin x cosn−1 x + (n − 1) cosn−2 x dx Z 1 n − 1 Z cosn x dx = sin x cosn−1 x + cosn−2 x dx n n

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