# Integral by Parts: Definite Integral

How to solve the definite integral of a product by using the integral by parts: formula, 2 examples, and their solutions.

## Formula

### Formula

The integral of a product, uv', can be solved
by using this formula.
ab uv' dx = [uv]ab - ∫ab u'v dx

### How to Choose u and v'

The priority of choosing u and v'
is the same as
the case of solving an indefinite integral.

For a product of two functions in an integral,
one is u, and the other is v'.

To choose u and v' properly,
remember this order of uv':

[u]
logarithmic
polynomial
trigonometric,
exponential
[v'].

Set u = (upper function) and v' = (lower function).

u is the function
whose integral is complex.

v' is the function
whose integral is simple.

## Example 1

### Solution

x sin x is the product of x and sin x.

So solve this
by using integral by parts.

The order of uv' is

[u]
logarithmic
polynomial (x)
trigonometric (sin x)
exponential
[v'].

So set
u = x and v' = sin x.

Write u = x.

Differentiate both sides.

Then u' = 1.

Derivative of a Polynomial

Write v' = sin x
next to u' = 1.

Integrate both sides.

Then v = -cos x.

Integral of sin x

Write this above v' = sin x.

u = x, v = -cos x
u' = 1

Then the given integral is equal to,
uv, x⋅(-cos x)
from 0 to π/3
minus
integral from 0 to π/3,
u'v, 1⋅(-cos x) dx.

x⋅(-cos x) = -x cos x

-∫0π/3 1⋅(-cos x) dx = +∫0π/3 cos x dx

Put 0 and π/3
into -x cos x.

Then -(π/3)⋅(cos π/3) - (0⋅cos 0).

Solve the integral.

Definite Integral: How to Solve

The integral of cos x is
sin x.

So -(π/3)⋅(cos π/3) - (0⋅cos 0) + [sin x]0π/3.

cos π/3 = 1/2
So -(π/3)⋅(cos π/3) = -(π/3)⋅(1/2).

Cosine Values of Commonly Used Angles

Put 0 and π/3
into sin x.

Then sin π/3 - sin 0.

-(π/3)⋅(1/2) = -π/6

sin π/3 = √3/2
sin 0 = 0

Sine Values of Commonly Used Angles

-π/6 + [√3/2 - 0] = √3/2 - π/6

So √3/2 - π/6 is the answer.

## Example 2

### Solution

(ln x)2 is the product of (ln x)2 and 1.

So solve this
by using integral by parts.

The order of uv' is

[u]
logarithmic [(ln x)2]
polynomial (1)
trigonometric
exponential
[v'].

So set
u = (ln x)2 and v' = 1.

Write u = (ln x)2.

Differentiate both sides.

Then u' = 2(ln x)⋅(1/x).

Derivative of a Composite Function

Derivative of ln x

Write v' = 1
next to u' = 2(ln x)1⋅(1/x).

Integrate both sides.

Then v = x.

Integral of a Polynomial

Write this above v' = 1.

u = (ln x)2, v = 2(ln x)⋅(1/x)
u' = x

Then the given integral is equal to,
uv, (ln x)2⋅x
from 1 to e
minus
integral from 1 to e,
u'v, 2(ln x)⋅(1/x)⋅x dx.

Put 1 and e
into (ln x)2⋅x.

Then (ln e)2⋅e - (ln 1)2⋅1.

Take 2 out from the integral.
And cancel (1/x) and x.

Then -∫1e 2(ln x)⋅(1/x)⋅x dx = -2⋅∫1e ln x dx

So (ln e)2⋅e - (ln 1)2⋅1 - 2⋅∫1e ln x dx.

(ln e)2⋅e = 12⋅e
-(ln 1)2⋅1 = -0⋅1 = 0.

Solve the integral.

The integral of ln x is
x ln x - x.

12⋅e = e

Put 1 and e
into x ln x - x.

Then -2[e ln e - e - (1 ln 1 - 1)].

e ln e = e⋅1

1 ln 1 = 1⋅0

e⋅1 - e = 0

(1⋅0 - 1) = (-1)

0 - (-1) = 0 + 1 = +1

-2[+1] = -2

So e - 2 is the answer.