CHAPTER 7 TECHNIQUES OF INTEGRATION 7.1 Integration by Parts
Integration by Parts, Part 2
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Transcript of Integration by Parts, Part 2
Example 2
Example 2
Let’s find the integral:
Example 2
Let’s find the integral: ∫x sin xdx
Example 2
Let’s find the integral: ∫x sin xdx
Let’s write again the formula of integration by parts:
Example 2
Let’s find the integral: ∫x sin xdx
Let’s write again the formula of integration by parts:∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
Example 2
Let’s find the integral: ∫x sin xdx
Let’s write again the formula of integration by parts:∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
In this case we choose:
Example 2
Let’s find the integral: ∫x sin xdx
Let’s write again the formula of integration by parts:∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
In this case we choose:
u(x) = x
Example 2
Let’s find the integral: ∫x sin xdx
Let’s write again the formula of integration by parts:∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
In this case we choose:
u(x) = x v ′(x) = sin x
Example 2
Let’s find the integral: ∫x sin xdx
Let’s write again the formula of integration by parts:∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
In this case we choose:
u(x) = x v ′(x) = sin xu′(x) = 1
Example 2
Let’s find the integral: ∫x sin xdx
Let’s write again the formula of integration by parts:∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
In this case we choose:
u(x) = x v ′(x) = sin xu′(x) = 1 v(x) = − cos x
Example 2
Let’s find the integral: ∫x sin xdx
Let’s write again the formula of integration by parts:∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
In this case we choose:
u(x) = x v ′(x) = sin xu′(x) = 1 v(x) = − cos x
So, we have that:
Example 2
Let’s find the integral: ∫x sin xdx
Let’s write again the formula of integration by parts:∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
In this case we choose:
u(x) = x v ′(x) = sin xu′(x) = 1 v(x) = − cos x
So, we have that: ∫x sin xdx
Example 2
Let’s find the integral: ∫x sin xdx
Let’s write again the formula of integration by parts:∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
In this case we choose:
u(x) = x v ′(x) = sin xu′(x) = 1 v(x) = − cos x
So, we have that: ∫��u(x)
x sin xdx
Example 2
Let’s find the integral: ∫x sin xdx
Let’s write again the formula of integration by parts:∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
In this case we choose:
u(x) = x v ′(x) = sin xu′(x) = 1 v(x) = − cos x
So, we have that: ∫x���:
v ′(x)sin xdx
Example 2
Let’s find the integral: ∫x sin xdx
Let’s write again the formula of integration by parts:∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
In this case we choose:
u(x) = x v ′(x) = sin xu′(x) = 1 v(x) = − cos x
So, we have that: ∫x sin xdx =
Example 2
Let’s find the integral: ∫x sin xdx
Let’s write again the formula of integration by parts:∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
In this case we choose:
u(x) = x v ′(x) = sin xu′(x) = 1 v(x) = − cos x
So, we have that: ∫x sin xdx = x
Example 2
Let’s find the integral: ∫x sin xdx
Let’s write again the formula of integration by parts:∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
In this case we choose:
u(x) = x v ′(x) = sin xu′(x) = 1 v(x) = − cos x
So, we have that: ∫x sin xdx = x(− cos x)
Example 2
Let’s find the integral: ∫x sin xdx
Let’s write again the formula of integration by parts:∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
In this case we choose:
u(x) = x v ′(x) = sin xu′(x) = 1 v(x) = − cos x
So, we have that: ∫x sin xdx =��
u(x)x(− cos x)
Example 2
Let’s find the integral: ∫x sin xdx
Let’s write again the formula of integration by parts:∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
In this case we choose:
u(x) = x v ′(x) = sin xu′(x) = 1 v(x) = − cos x
So, we have that: ∫x sin xdx = x���
��:v(x)
(− cos x)
Example 2
Let’s find the integral: ∫x sin xdx
Let’s write again the formula of integration by parts:∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
In this case we choose:
u(x) = x v ′(x) = sin xu′(x) = 1 v(x) = − cos x
So, we have that: ∫x sin xdx = x(− cos x)−
Example 2
Let’s find the integral: ∫x sin xdx
Let’s write again the formula of integration by parts:∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
In this case we choose:
u(x) = x v ′(x) = sin xu′(x) = 1 v(x) = − cos x
So, we have that:∫x sin xdx = x(− cos x) −
∫1. (− cos x) dx
Example 2
Let’s find the integral: ∫x sin xdx
Let’s write again the formula of integration by parts:∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
In this case we choose:
u(x) = x v ′(x) = sin xu′(x) = 1 v(x) = − cos x
So, we have that:
∫x sin xdx = x(− cos x) −
∫���
u′(x)
1. (− cos x) dx
Example 2
Let’s find the integral: ∫x sin xdx
Let’s write again the formula of integration by parts:∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
In this case we choose:
u(x) = x v ′(x) = sin xu′(x) = 1 v(x) = − cos x
So, we have that:∫x sin xdx = x(− cos x) −
∫1.���
��:v(x)
(− cos x)dx
Example 2
Example 2
So our integral becomes:
Example 2
So our integral becomes:∫x sin xdx = −x cos x +
∫cos xdx
Example 2
So our integral becomes:∫x sin xdx = −x cos x +
∫cos xdx
And that’s easy to solve:
Example 2
So our integral becomes:∫x sin xdx = −x cos x +
∫cos xdx
And that’s easy to solve:∫x sin xdx = −x cos x + sin x
Example 2
So our integral becomes:∫x sin xdx = −x cos x +
∫cos xdx
And that’s easy to solve:∫x sin xdx = −x cos x + sin x
Trick Number 1
Trick Number 1
Take v ′(x) = 1.
Trick Number 1
Take v ′(x) = 1. For example:
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1u′(x) = 1
x
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1u′(x) = 1
x v(x) = x
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1u′(x) = 1
x v(x) = x
So we have:
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1u′(x) = 1
x v(x) = x
So we have: ∫ln xdx
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1u′(x) = 1
x v(x) = x
So we have: ∫ln xdx =
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1u′(x) = 1
x v(x) = x
So we have: ∫ln xdx =
∫ln x .1dx
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1u′(x) = 1
x v(x) = x
So we have: ∫ln xdx =
∫���*
u(x)ln x .1dx
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1u′(x) = 1
x v(x) = x
So we have:
∫ln xdx =
∫ln x .���
v ′(x)
1dx
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1u′(x) = 1
x v(x) = x
So we have:
∫ln xdx =
∫ln x .���
v ′(x)
1dx
= ln x
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1u′(x) = 1
x v(x) = x
So we have:
∫ln xdx =
∫ln x .���
v ′(x)
1dx
= ln x .x
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1u′(x) = 1
x v(x) = x
So we have:
∫ln xdx =
∫ln x .���
v ′(x)
1dx
=���*
u(x)ln x .x
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1u′(x) = 1
x v(x) = x
So we have:
∫ln xdx =
∫ln x .���
v ′(x)
1dx
= ln x .��v(x)
x
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1u′(x) = 1
x v(x) = x
So we have:
∫ln xdx =
∫ln x .���
v ′(x)
1dx
= ln x .x−
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1u′(x) = 1
x v(x) = x
So we have:
∫ln xdx =
∫ln x .���
v ′(x)
1dx
= ln x .x −∫
1
x.xdx
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1u′(x) = 1
x v(x) = x
So we have:
∫ln xdx =
∫ln x .���
v ′(x)
1dx
= ln x .x −∫����
u′(x)
1
x.xdx
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1u′(x) = 1
x v(x) = x
So we have:
∫ln xdx =
∫ln x .���
v ′(x)
1dx
= ln x .x −∫
1
x.��
v(x)xdx
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1u′(x) = 1
x v(x) = x
So we have:
∫ln xdx =
∫ln x .���
v ′(x)
1dx
= ln x .x −∫
1
x.xdx
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1u′(x) = 1
x v(x) = x
So we have:
∫ln xdx =
∫ln x .���
v ′(x)
1dx
= ln x .x −∫
1
�x.�xdx
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1u′(x) = 1
x v(x) = x
So we have:
∫ln xdx =
∫ln x .���
v ′(x)
1dx
= ln x .x −∫
1
�x.�xdx = x ln x −
∫dx
Trick Number 1
Take v ′(x) = 1. For example:∫ln xdx
We take:
u(x) = ln x v ′(x) = 1u′(x) = 1
x v(x) = x
So we have:
∫ln xdx =
∫ln x .���
v ′(x)
1dx
= ln x .x −∫
1
�x.�xdx = x ln x −
∫dx = x ln x − x
Trick Number 1
Trick Number 1
So we have:
Trick Number 1
So we have: ∫ln xdx = x (ln x − 1)
Trick Number 1
So we have: ∫ln xdx = x (ln x − 1) + C