Math Calculators, Lessons and Formulas

It is time to solve your math problem

mathportal.org
Integration Techniques: (lesson 2 of 4)

## Integration by Parts

Theorem:

The formula for the method of integration by parts is:

$$\color{blue}{\int udv = u \cdot v - \int vdu}$$

There are four steps how to use this formula:

Step 1: Identify $u$ and $dv$. Priorities for choosing $u$ are: 1. $u = lnx$ 2. $u = x^n$ 3. $u = e^{ax}$

Step 2: Compute $du$ and $v$

Step 3: Use the formula for the integration by parts

Example 1: Evaluate the following integral

$$\int x \cdot \sin x dx$$

Solution:

Step 1: In this example we choose $\color{blue}{u = x}$ and $\color{red}{dv}$ will be everything else that remains.

\begin{aligned} &\int {\color{blue}{x}} \cdot \color{red}{\sin xdx} \\ &\color{blue}{u = x} \ \ \color{red}{dv = \sin xdx} \end{aligned}

Step 2: Compute $du$ and $v$

\begin{aligned} \int x \cdot \sin x \, dx & \\ \color{blue}{u = x} \ \ & \ \ \color{red}{dv = \sin xdx}\\ \color{blue}{du = x'dx} \ \ & \ \ \color{red}{v = \int \sin xdx} \\ \color{blue}{du = dx} \ \ & \ \ \color{red}{v = - \cos x} \end{aligned}

Step 3: Apply the formula.

$$\int x \cdot \sin x dx = \color{blue}{x} \cdot \color{red}{(-\cos x)} - \int \color{red}{(- \cos x)} dx$$

Therefore:

$\int x \cdot \sin xdx = x \cdot (-\cos x) - \int (-\cos x) dx = -x \cdot \cos x + \int \cos xdx = -x \cdot \cos x + \sin x + C$

Example 2: Evaluate the following integral

$$\int x \cdot \ln xdx$$

Solution:

Step 1: In this example, choose $\color{blue}{u = ln x}$ (first priority) and $\color{red}{dv = x dx}$. \begin{aligned} &\int \color{red}{x} \cdot \color{blue}{\ln x} \color{red}{dx} \\ &\color{blue}{u = \ln x} \ \ \color{red}{dv = xdx} \end{aligned}

Step 2: Compute $du$ and $v$

\begin{aligned} &\int x \cdot \ln xdx \\ &u = \ln x \ \ \ \ dv = xdx \\ &du = (\ln x)' dx \ \ \ \ v = \int xdx \\ &\color{blue}{du = \frac{1}{x} dx} \ \ \ \ \color{red}{v = \frac{x^2}{2}} \end{aligned}

Step 3: Use the formula.

\begin{aligned} &\int x \cdot \ln xdx = \color{blue}{\ln x} \cdot \color{red}{\frac{x^2}{2}} - \int \frac{x^2}{2} \cdot \color{purple}{\frac{1}{x}dx} \\ &\color{blue}{u = \ln x} \ \ \ \ dv = xdx \\ &du = (\ln x)' dx \ \ \ \ v = \int xdx \\ &\color{purple}{du = \frac{1}{x} dx} \ \ \ \ \color{red}{v = \frac{x^2}{2}} \end{aligned}

The solution is:

$$\int x \cdot \ln xdx = \ln x \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{1}{2} x^2 \ln x - \frac{1}{2} \int xdx = \frac{1}{2} x^2 \ln x - \frac{x^2}{4} + C$$

Exercise 1: Evaluate the following integrals

Level 1

 $$\color{blue}{\int x \cdot \cos xdx = }$$ $x \cdot \cos x - \sin x + C$ $x \cdot \cos x + \sin x + C$ $x \cdot \sin x - \cos x + C$ $x \cdot \sin x + \cos x + C$

Level 2

 $$\color{blue}{\int x^2 \ln xdx = }$$ $\frac{1}{3}x^3 \cdot \ln x - \frac{1}{9}x^3 + C$ $\frac{1}{3}x^3 \cdot \ln x + \frac{1}{9}x^3 + C$ $\frac{1}{3}x^3 \cdot \ln x + \frac{1}{3}x^3 + C$ $\frac{1}{9}x^3 \cdot \ln x + \frac{1}{3}x^3 + C$

### Integration by parts twice

Example 3: Evaluate the following integral

$$\int x^2 \cdot e^x dx$$

Solution:

Let:

$$u = x^2 \ \ \ \ dv = e^x dx$$

So that

\begin{aligned} &u = x^2 \ \ \ \ dv = e^x dx \\ &u = (x^2)' dx \ \ \ \ v = \int e^x dx \\ &u = 2xdx \ \ \ \ v = e^x \end{aligned}

Therefore:

$$\int x^2 \cdot e^x dx = 2x \cdot e^x - \int 2xe^x dx = 2xe^x - 2 \int xe^x dx$$

It is needed to perform integration by parts again:

\begin{aligned} \int x^2 \cdot e^x dx &= x^2 \cdot e^x - \int 2xe^x dx = x^2 e^x - 2 \color{blue}{\int x e^x dx} = \\ &= x^2 e^x - 2 \color{blue}{(x \cdot e^x - \int e^x dx)} = x^2 e^x - 2 (xe^x - e^x) + C = \\ &= x^2 e^x - 2xe^x + 2^x + C \end{aligned}
 Try yourself $$\color{blue}{\int x^2 \sin xdx =}$$ $x^2 \cos x - 2x \sin x + \\ + 2 \cos x + C$ $x^2 \sin x - 2x \cos x + \\ + 2 \sin x + C$ $-x^2 \cos x + 2x \sin x + \\ + 2 \cos x + C$ $x^2 \sin x - 2x \cos x + \\ - 2 \sin x + C$