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« Integration by Substitution
Integrals Involving Trigonometric Functions »
Integration Techniques: (lesson 2 of 4)

Integration by Parts

Theorem:

The formula for the method of integration by parts is:

∫udv=u⋅v−∫vdu \color{blue}{\int udv = u \cdot v - \int vdu}

There are four steps how to use this formula:

Step 1: Identify uu and dvdv. Priorities for choosing uu are: 1. u=lnxu = lnx 2. u=xnu = x^n 3. u=eaxu = e^{ax}

Step 2: Compute dudu and vv

Step 3: Use the formula for the integration by parts

Example 1: Evaluate the following integral

∫x⋅sin⁡xdx\int x \cdot \sin x dx

Solution:

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

∫x⋅sin⁡xdxu=x  dv=sin⁡xdx \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 dudu and vv

∫x⋅sin⁡x dxu=x    dv=sin⁡xdxdu=x′dx    v=∫sin⁡xdxdu=dx    v=−cos⁡x \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.

∫x⋅sin⁡xdx=x⋅(−cos⁡x)−∫(−cos⁡x)dx \int x \cdot \sin x dx = \color{blue}{x} \cdot \color{red}{(-\cos x)} - \int \color{red}{(- \cos x)} dx

Therefore:

∫x⋅sin⁡xdx=x⋅(−cos⁡x)−∫(−cos⁡x)dx=−x⋅cos⁡x+∫cos⁡xdx=−x⋅cos⁡x+sin⁡x+C\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

∫x⋅ln⁡xdx\int x \cdot \ln xdx

Solution:

Step 1: In this example, choose u=lnx \color{blue}{u = ln x} (first priority) and dv=xdx\color{red}{dv = x dx}. ∫x⋅ln⁡xdxu=ln⁡x  dv=xdx \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 dudu and vv

∫x⋅ln⁡xdxu=ln⁡x    dv=xdxdu=(ln⁡x)′dx    v=∫xdxdu=1xdx    v=x22 \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.

∫x⋅ln⁡xdx=ln⁡x⋅x22−∫x22⋅1xdxu=ln⁡x    dv=xdxdu=(ln⁡x)′dx    v=∫xdxdu=1xdx    v=x22 \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:

∫x⋅ln⁡xdx=ln⁡x⋅x22−∫x22⋅1xdx=12x2ln⁡x−12∫xdx=12x2ln⁡x−x24+C \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

∫x⋅cos⁡xdx= \color{blue}{\int x \cdot \cos xdx = } x⋅cos⁡x−sin⁡x+C x \cdot \cos x - \sin x + C
x⋅cos⁡x+sin⁡x+C x \cdot \cos x + \sin x + C
x⋅sin⁡x−cos⁡x+C x \cdot \sin x - \cos x + C
x⋅sin⁡x+cos⁡x+C x \cdot \sin x + \cos x + C

Level 2

∫x2ln⁡xdx= \color{blue}{\int x^2 \ln xdx = } 13x3⋅ln⁡x−19x3+C \frac{1}{3}x^3 \cdot \ln x - \frac{1}{9}x^3 + C
13x3⋅ln⁡x+19x3+C \frac{1}{3}x^3 \cdot \ln x + \frac{1}{9}x^3 + C
13x3⋅ln⁡x+13x3+C \frac{1}{3}x^3 \cdot \ln x + \frac{1}{3}x^3 + C
19x3⋅ln⁡x+13x3+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

∫x2⋅exdx\int x^2 \cdot e^x dx

Solution:

Let:

u=x2    dv=exdx u = x^2 \ \ \ \ dv = e^x dx

So that

u=x2    dv=exdxu=(x2)′dx    v=∫exdxu=2xdx    v=ex \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:

∫x2⋅exdx=2x⋅ex−∫2xexdx=2xex−2∫xexdx \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:

∫x2⋅exdx=x2⋅ex−∫2xexdx=x2ex−2∫xexdx==x2ex−2(x⋅ex−∫exdx)=x2ex−2(xex−ex)+C==x2ex−2xex+2x+C \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
∫x2sin⁡xdx= \color{blue}{\int x^2 \sin xdx =}
x2cos⁡x−2xsin⁡x++2cos⁡x+C x^2 \cos x - 2x \sin x + \\ + 2 \cos x + C x2sin⁡x−2xcos⁡x++2sin⁡x+C x^2 \sin x - 2x \cos x + \\ + 2 \sin x + C −x2cos⁡x+2xsin⁡x++2cos⁡x+C -x^2 \cos x + 2x \sin x + \\ + 2 \cos x + C x2sin⁡x−2xcos⁡x+−2sin⁡x+C x^2 \sin x - 2x \cos x + \\ - 2 \sin x + C

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Paul Erdos

Random Quote

A mathematician is a device for turning coffee into theorems.

Paul Erdos