7. Integration by Parts

by M. Bourne

Sometimes we meet an integration that is the product of 2 functions. We may be able to integrate such products by using Integration by Parts.

If u and v are functions of x, the product rule for differentiation that we met earlier gives us:

`d/(dx)(uv)=u(dv)/(dx)+v(du)/(dx)`

Rearranging, we have:

`u(dv)/(dx)=d/(dx)(uv)-v(du)/(dx)`

Integrating throughout with respect to x, we obtain the formula for integration by parts:

Formula for integration by parts

This formula allows us to turn a complicated integral into more simple ones. We must make sure we choose u and dv carefully.

NOTE: The function u is chosen so that `(du)/(dx)` is simpler than u.

Continues below

Priorities for Choosing u

When you have a mix of functions in the expression to be integrated, use the following for your choice of `u`, in order.

1. Let `u = ln x`

2. Let `u = x^n`

3. Let `u = e^(nx)`

Example 1

`intx\ sin 2x\ dx`

Solution

We need to choose `u`. In this question we don't have any of the functions suggested in the "priorities" list above.

We could let `u = x` or `u = sin 2x`, but usually only one of them will work. In general, we choose the one that allows `(du)/(dx)` to be of a simpler form than u.

So for this example, we choose u = x and so `dv` will be the "rest" of the integral, dv = sin 2x dx.

We have `u = x` so `du = dx`.

Also `dv = sin 2x\ dx` and integrating gives:

`v=intsin 2x\ dx`

`=(-cos 2x)/2`

Substituting these 4 expressions into the integration by parts formula, we get (using color-coding so it's easier to see where things come from):

`int \color{green}{\underbrace{u}}\ \ \ \color{red}{\underbrace{dv}}\ \ ` ` =\ \ \color{green}{\underbrace{u}}\ \ \ \color{blue}{\underbrace{v}} \ \ -\ \ int \color{blue}{\underbrace{v}}\ \ \color{magenta}{\underbrace{du}}`

`int \color{green}{\fbox{:x:}}\ \color{red}{\fbox{:sin 2x dx:}} = \color{green}{\fbox{:x:}}\ \color{blue}{\fbox{:{-cos2x}/2:}} - int \color{blue}{\fbox{:{-cos2x}/2:}\ \color{magenta}{\fbox{:dx:}}`

` = (-xcos2x)/2 + 1/2 int cos2x dx`

` = (-xcos2x)/2 + 1/2 (sin2x)/2 +K`

` = (-xcos2x)/2 + (sin2x)/4 +K`

If the above is a little hard to follow (because of the line breaks), here it is again in a different format:

Integration by parts example steps

Example 2

`int x sqrt(x+1) dx`

Answer

`intxsqrt(x+1)\ dx`

We could let `u=x` or `u=sqrt(x+1)`.

Once again, we choose the one that allows `(du)/(dx)` to be of a simpler form than `u`, so we choose `u=x`.

Therefore `du = dx`. With this choice, `dv` must be the "rest" of the integral: `dv=sqrt(x+1)\ dx`.

`u = x` so `du=dx`.

`dv=sqrt(x+1)\ dx`, and integrating gives:

`v=intsqrt(x+1) dx`

`=int(x+1)^(1//2)dx`

`=2/3(x+1)^(3//2)`

Substituting into the integration by parts formula, we get:

`int \color{green}{\underbrace{u}}\ \ \ \color{red}{\underbrace{dv}}\ \ ` ` =\ \ \color{green}{\underbrace{u}}\ \ \ \color{blue}{\underbrace{v}} \ \ -\ \ int \color{blue}{\underbrace{v}}\ \ \color{magenta}{\underbrace{du}}`

`int \color{green}{\fbox{:x:}}\ \color{red}{\fbox{:sqrt(x+1) dx:}} = \color{green}{\fbox{:x:}}\ \color{blue}{\fbox{:2/3(x+1)^(3//2):}} ` `- int \color{blue}{\fbox{:2/3(x+1)^(3//2):}\ \color{magenta}{\fbox{:dx:}}`

` = (2x)/3(x+1)^(3//2) - 2/3 int (x+1)^{3//2}dx`

` = (2x)/3(x+1)^(3//2) ` `- 2/3(2/5) (x+1)^{5//2} +K`

` = (2x)/3(x+1)^(3//2)- 4/15(x+1)^{5//2} +K`

Once again, here it is again in a different format:

Integration by parts example

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Example 3

`intx^2 ln 4x\ dx`

Answer

`intx^2ln\ 4x\ dx`

We could let `u=x^2` or `u=ln\ 4x`..

Considering the priorities given above, we choose `u = ln\ 4x` and so `dv` will be the rest of the expression to be integrated `dv = x^2\ dx`.

With `u=ln\ 4x`, we have `du=dx/x`.

Integrating `dv = x^2\ dx` gives:

`v=intx^2dx=x^3/3`

Substituting in the Integration by Parts formula, we get:

`int \color{green}{\underbrace{u}}\ \ \ \color{red}{\underbrace{dv}}\ \ ` ` =\ \ \color{green}{\underbrace{u}}\ \ \ \color{blue}{\underbrace{v}} \ \ -\ \ int \color{blue}{\underbrace{v}}\ \ \color{magenta}{\underbrace{du}}`

`int \color{green}{\fbox{:x^2:}}\ \color{red}{\fbox{:ln 4x dx:}} = \color{green}{\fbox{:ln 4x:}}\ \color{blue}{\fbox{:x^3/3:}} ` `- int \color{blue}{\fbox{:x^3/3:}\ \color{magenta}{\fbox{:dx/x:}}`

` = (x^3 ln 4x)/3 - 1/3 int x^2 dx`

` = (x^3 ln 4x)/3 ` `- 1/3 x^3/3 +K`

` = (x^3 ln 4x)/3 - x^3/9 +K`

Once again, here it is again in a different format:

Integration by parts example

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Example 4

`intx\ sec^2 x\ dx`

Answer

`int x\ sec^2 x\ dx`

We choose `u=x` (since it will give us a simpler `du`) and this gives us `du=dx`.

Then `dv` will be `dv=sec^2x\ dx` and integrating this gives `v=tan x`.

Substituting these into the Integration by Parts formula gives:

`int x\ sec^2 x\ dx =intu\ dv`

`=uv-intv\ du`

`=(x)(tan x)-int(tan x)dx`

`=x\ tan x-(-ln\ |cos x|)+K`

`=x\ tan x+ln\ |cos x|+K`

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Example 5

`intx^2e^(-x)dx`

Answer

`intx^2 e^-x dx`

The 2nd and 3rd "priorities" for choosing `u` given earlier said:

2. Let `u = x^n`

3. Let `u = e^(nx)`

This questions has both a power of `x` and an exponential expression. But we choose `u=x^2` as it has a higher priority than the exponential. (You could try it the other way round, with `u=e^-x` to see for yourself why it doesn't work.)

So `u=x^2` and this gives `du=2x\ dx`.

That leaves `dv=e^-x\ dx` and integrating this gives us `v=-e^-x`.

We substitute these into the Integration by Parts formula to give:

`intx^2 e^-x dx =intu\ dv`

`=uv-intv\ du`

`=x^2(-e^-x)-int(-e^-x)(2x\ dx) `

`=-x^2e^-x+2intxe^-x dx `

Now, the integral we are left with cannot be found immediately. We need to perform integration by parts again, for this new integral.

This time we choose `u=x` giving `du=dx`.

Once again we will have `dv=e^-x\ dx` and integrating this gives us `v=-e^-x`.

Substituting into the integration by parts formula gives:

`int x e^-x dx =intu\ dv`

`=uv-intv\ du`

`=x(-e^-x)-int(-e^-x)dx`

`=-xe^-x+inte^-x dx`

`=-xe^-x-e^-x `

So putting this answer together with the answer for the first part, we have the final solution:

`intx^2e^-xdx =-x^2e^-x+2(-xe^-x-e^-x) `

`=-e^-x(x^2+2x+2)+K`

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Example 6

`int ln x dx`

Answer

`int ln\ x\ dx`

Our priorities list above tells us to choose the logarithm expression for `u`. (of course, there's no other choice here. :-)

So with `u=ln\ x`, we have `du=dx/x`.

Then `dv` will simply be `dv=dx` and integrating this gives `v=x`.

Subsituting these into the Integration by Parts formula gives:

`int ln\ x\ dx=int u\ dv`

`=uv-intv\ du`

`=x\ ln\ x-intx(dx)/x`

`=x\ ln\ x-intdx`

`=x\ ln\ x-x+K`

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Example 7

`intarcsin x dx`

Answer

Using integration by parts, we set:

`u=arcsin x`, giving `du=1/sqrt(1-x^2)dx`.

Then `dv=dx` and integrating gives us `v=x`.

We now use:

`intu\ dv=uv-intv\ du`

This gives us:

`int arcsin x\ dx` `=x\ arcsin x-intx/(sqrt(1-x^2))dx`

Now, for that remaining integral, we just use a substitution (I'll use `p` for the substitution since we are using `u` in this question already):

`p = 1 − x^2`

So `dp=-2x\ dx`

This will yield:

`intx/(sqrt(1-x^2))dx =-1/2int(dp)/sqrtp`

`=-1/2(2sqrtp)+K`

`=-sqrt(1-x^2)+K`

So our final answer is:

`int arcsin x\ dx =x\ arcsin x-(-sqrt(1-x^2))+K `

`= \x\ arcsin x+sqrt(1-x^2)+K`

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This time we integrated an inverse trigonometric function (as opposed to the earlier type where we obtained inverse trigonometric functions in our answer). See Integration: Inverse Trigonometric Forms.

Alternate Method for Integration by Parts

Here's an alternative method for problems that can be done using Integration by Parts. You may find it easier to follow.

Tanzalin Method