# 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

The integral of a product, uv', can be solved

by using this formula.

∫_{a}^{b} uv' dx = [uv]_{a}^{b} - ∫_{a}^{b} 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∫_{0}^{π/3} (x sin x) dx

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

-∫_{0}^{π/3} 1⋅(-cos x) dx = +∫_{0}^{π/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]_{0}^{π/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∫_{1}^{e} (ln x)^{2} dx

(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 -∫_{1}^{e} 2(ln x)⋅(1/x)⋅x dx = -2⋅∫_{1}^{e} ln x dx

So (ln e)^{2}⋅e - (ln 1)^{2}⋅1 - 2⋅∫_{1}^{e} ln x dx.

(ln e)^{2}⋅e = 1^{2}⋅e

-(ln 1)^{2}⋅1 = -0⋅1 = 0.

Solve the integral.

The integral of ln x is

x ln x - x.

1^{2}⋅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.