Integration by Parts, Part 2

Example 2

Let’s find the integral:

Example 2

Let’s find the integral: ∫x sin xdx

Example 2

Let’s write again the formula of integration by parts:

Example 2

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

∫u′(x)v(x)dx

In this case we choose:

Example 2

∫u′(x)v(x)dx

u(x) = x

Example 2

∫u′(x)v(x)dx

u(x) = x v ′(x) = sin x

Example 2

∫u′(x)v(x)dx

u(x) = x v ′(x) = sin xu′(x) = 1

Example 2

∫u′(x)v(x)dx

u(x) = x v ′(x) = sin xu′(x) = 1 v(x) = − cos x

Example 2

∫u′(x)v(x)dx

So, we have that:

Example 2

∫u′(x)v(x)dx

So, we have that: ∫x sin xdx

Example 2

∫u′(x)v(x)dx

So, we have that: ∫��u(x)

x sin xdx

Example 2

∫u′(x)v(x)dx

So, we have that: ∫x��:

v ′(x)sin xdx

Example 2

∫u′(x)v(x)dx

So, we have that: ∫x sin xdx =

Example 2

∫u′(x)v(x)dx

So, we have that: ∫x sin xdx = x

Example 2

∫u′(x)v(x)dx

So, we have that: ∫x sin xdx = x(− cos x)

Example 2

∫u′(x)v(x)dx

So, we have that: ∫x sin xdx =��

u(x)x(− cos x)

Example 2

∫u′(x)v(x)dx

So, we have that: ∫x sin xdx = x��

��:v(x)

(− cos x)

Example 2

∫u′(x)v(x)dx

So, we have that: ∫x sin xdx = x(− cos x)−

Example 2

∫u′(x)v(x)dx

So, we have that:∫x sin xdx = x(− cos x) −

∫1. (− cos x) dx

Example 2

∫u′(x)v(x)dx

So, we have that:

∫x sin xdx = x(− cos x) −

∫��

u′(x)

1. (− cos x) dx

Example 2

∫u′(x)v(x)dx

So, we have that:∫x sin xdx = x(− cos x) −

∫1.��

��:v(x)

(− cos x)dx

Example 2

So our integral becomes:

Example 2

So our integral becomes:∫x sin xdx = −x cos x +

∫cos xdx

Example 2

∫cos xdx

And that’s easy to solve:

Example 2

∫cos xdx

And that’s easy to solve:∫x sin xdx = −x cos x + sin x

Example 2

∫cos xdx

And that’s easy to solve:∫x sin xdx = −x cos x + sin x

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

We take:

Trick Number 1

We take:

u(x) = ln x

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1u′(x) = 1

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1u′(x) = 1

x v(x) = x

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1u′(x) = 1

x v(x) = x

So we have:

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1u′(x) = 1

x v(x) = x

So we have: ∫ln xdx

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1u′(x) = 1

x v(x) = x

So we have: ∫ln xdx =

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1u′(x) = 1

x v(x) = x

∫ln x .1dx

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1u′(x) = 1

x v(x) = x

∫��*

u(x)ln x .1dx

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1u′(x) = 1

x v(x) = x

So we have:

∫ln xdx =

∫ln x .��

v ′(x)

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1u′(x) = 1

x v(x) = x

So we have:

∫ln xdx =

∫ln x .��

v ′(x)

= ln x

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1u′(x) = 1

x v(x) = x

So we have:

∫ln xdx =

∫ln x .��

v ′(x)

= ln x .x

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1u′(x) = 1

x v(x) = x

So we have:

∫ln xdx =

∫ln x .��

v ′(x)

=��*

u(x)ln x .x

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1u′(x) = 1

x v(x) = x

So we have:

∫ln xdx =

∫ln x .��

v ′(x)

= ln x .��v(x)

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1u′(x) = 1

x v(x) = x

So we have:

∫ln xdx =

∫ln x .��

v ′(x)

= ln x .x−

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1u′(x) = 1

x v(x) = x

So we have:

∫ln xdx =

∫ln x .��

v ′(x)

= ln x .x −∫

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1u′(x) = 1

x v(x) = x

So we have:

∫ln xdx =

∫ln x .��

v ′(x)

= ln x .x −∫��

u′(x)

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1u′(x) = 1

x v(x) = x

So we have:

∫ln xdx =

∫ln x .��

v ′(x)

= ln x .x −∫

x.��

v(x)xdx

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1u′(x) = 1

x v(x) = x

So we have:

∫ln xdx =

∫ln x .��

v ′(x)

= ln x .x −∫

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1u′(x) = 1

x v(x) = x

So we have:

∫ln xdx =

∫ln x .��

v ′(x)

= ln x .x −∫

�x.�xdx

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1u′(x) = 1

x v(x) = x

So we have:

∫ln xdx =

∫ln x .��

v ′(x)

= ln x .x −∫

�x.�xdx = x ln x −

Trick Number 1

We take:

u(x) = ln x v ′(x) = 1u′(x) = 1

x v(x) = x

So we have:

∫ln xdx =

∫ln x .��

v ′(x)

= ln x .x −∫

�x.�xdx = x ln x −

∫dx = x ln x − x

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

Integration by Parts, Part 2

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