Integration by Parts

The product rule in reverse for integrating products of functions

🔀 Integration by Parts

What is Integration by Parts?

Integration by parts is the reverse of the product rule for derivatives. It's used to integrate products of functions.

💡 Key Idea: Trade one integral for another (hopefully easier) integral!


The Product Rule (Review)

The product rule for derivatives states:

ddx[u(x)v(x)]=u(x)v(x)+u(x)v(x)\frac{d}{dx}[u(x) \cdot v(x)] = u'(x)v(x) + u(x)v'(x)


Deriving the Formula

Integrate both sides of the product rule:

ddx[uv]dx=uvdx+uvdx\int \frac{d}{dx}[uv]\,dx = \int u'v\,dx + \int uv'\,dx

uv=uvdx+uvdxuv = \int u'v\,dx + \int uv'\,dx

Rearranging:

uvdx=uvuvdx\int uv'\,dx = uv - \int u'v\,dx


The Integration by Parts Formula

udv=uvvdu\int u\,dv = uv - \int v\,du

Or equivalently:

u(x)v(x)dx=u(x)v(x)u(x)v(x)dx\int u(x)v'(x)\,dx = u(x)v(x) - \int u'(x)v(x)\,dx


How to Apply Integration by Parts

Step-by-Step Process

Step 1: Identify uu and dvdv from the integrand

  • Choose what to call uu
  • The rest becomes dvdv

Step 2: Differentiate uu to get dudu

Step 3: Integrate dvdv to get vv

Step 4: Apply the formula: udv=uvvdu\int u\,dv = uv - \int v\,du

Step 5: Evaluate the new integral vdu\int v\,du

Step 6: Simplify and add +C+C


Choosing uu and dvdv

The LIATE Rule (or ILATE)

Choose uu in this priority order:

L - Logarithmic functions (lnx\ln x, logx\log x)

I - Inverse trig functions (arctanx\arctan x, arcsinx\arcsin x)

A - Algebraic functions (xx, x2x^2, x\sqrt{x})

T - Trigonometric functions (sinx\sin x, cosx\cos x, tanx\tan x)

E - Exponential functions (exe^x, axa^x)

Pick the one that comes first for uu, the rest becomes dvdv!


Example 1: Basic Application

Evaluate xexdx\int x e^x\,dx

Step 1: Choose uu and dvdv

Using LIATE:

  • Algebraic (xx) comes before Exponential (exe^x)

Let u=xu = x (algebraic)

Let dv=exdxdv = e^x\,dx (exponential)


Step 2: Find dudu

du=dxdu = dx


Step 3: Find vv

v=exdx=exv = \int e^x\,dx = e^x

(Don't add +C to vv!)


Step 4: Apply formula

xexdx=uvvdu\int x e^x\,dx = uv - \int v\,du

=xexexdx= x \cdot e^x - \int e^x \cdot dx

=xexex+C= xe^x - e^x + C

=ex(x1)+C= e^x(x-1) + C


Check: ddx[ex(x1)]=ex(x1)+ex(1)=ex(x1+1)=xex\frac{d}{dx}[e^x(x-1)] = e^x(x-1) + e^x(1) = e^x(x-1+1) = xe^x


Example 2: With Logarithm

Evaluate lnxdx\int \ln x\,dx

Trick: Write as 1lnxdx\int 1 \cdot \ln x\,dx


Step 1: Choose uu and dvdv

Using LIATE:

  • Logarithmic (lnx\ln x) comes first

Let u=lnxu = \ln x

Let dv=1dxdv = 1\,dx (or just dxdx)


Step 2: Find dudu

du=1xdxdu = \frac{1}{x}\,dx


Step 3: Find vv

v=1dx=xv = \int 1\,dx = x


Step 4: Apply formula

lnxdx=uvvdu\int \ln x\,dx = uv - \int v\,du

=(lnx)(x)x1xdx= (\ln x)(x) - \int x \cdot \frac{1}{x}\,dx

=xlnx1dx= x\ln x - \int 1\,dx

=xlnxx+C= x\ln x - x + C

=x(lnx1)+C= x(\ln x - 1) + C


Example 3: Trigonometric

Evaluate xsinxdx\int x\sin x\,dx

Step 1: Choose uu and dvdv

Using LIATE:

  • Algebraic (xx) before Trigonometric (sinx\sin x)

Let u=xu = x

Let dv=sinxdxdv = \sin x\,dx


Step 2: Find dudu

du=dxdu = dx


Step 3: Find vv

v=sinxdx=cosxv = \int \sin x\,dx = -\cos x


Step 4: Apply formula

xsinxdx=uvvdu\int x\sin x\,dx = uv - \int v\,du

=x(cosx)(cosx)dx= x(-\cos x) - \int (-\cos x)\,dx

=xcosx+cosxdx= -x\cos x + \int \cos x\,dx

=xcosx+sinx+C= -x\cos x + \sin x + C


Repeated Integration by Parts

Sometimes you need to apply integration by parts multiple times.

Example 4: Apply Twice

Evaluate x2exdx\int x^2 e^x\,dx

First application:

u=x2u = x^2, dv=exdxdv = e^x\,dx

du=2xdxdu = 2x\,dx, v=exv = e^x

x2exdx=x2exex2xdx\int x^2 e^x\,dx = x^2 e^x - \int e^x \cdot 2x\,dx

=x2ex2xexdx= x^2 e^x - 2\int xe^x\,dx


Second application on xexdx\int xe^x\,dx:

u=xu = x, dv=exdxdv = e^x\,dx

du=dxdu = dx, v=exv = e^x

xexdx=xexexdx=xexex\int xe^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x


Combine:

x2exdx=x2ex2(xexex)+C\int x^2 e^x\,dx = x^2 e^x - 2(xe^x - e^x) + C

=x2ex2xex+2ex+C= x^2 e^x - 2xe^x + 2e^x + C

=ex(x22x+2)+C= e^x(x^2 - 2x + 2) + C


Tabular Method (For Repeated Parts)

When you need to apply parts multiple times with polynomial times exponential/trig:

Example: x3exdx\int x^3 e^x\,dx

| Sign | uu and derivatives | dvdv and integrals | |------|-------------------|-------------------| | + | x3x^3 | exe^x | | − | 3x23x^2 | exe^x | | + | 6x6x | exe^x | | − | 66 | exe^x | | + | 00 | exe^x |

Answer: x3ex3x2ex+6xex6ex+C=ex(x33x2+6x6)+Cx^3e^x - 3x^2e^x + 6xe^x - 6e^x + C = e^x(x^3 - 3x^2 + 6x - 6) + C

Multiply diagonally with alternating signs!


Circular Integration by Parts

Sometimes integration by parts leads back to the original integral!

Example 5: Circular Case

Evaluate exsinxdx\int e^x \sin x\,dx

First application:

u=sinxu = \sin x, dv=exdxdv = e^x\,dx

du=cosxdxdu = \cos x\,dx, v=exv = e^x

exsinxdx=exsinxexcosxdx\int e^x \sin x\,dx = e^x \sin x - \int e^x \cos x\,dx


Second application on excosxdx\int e^x \cos x\,dx:

u=cosxu = \cos x, dv=exdxdv = e^x\,dx

du=sinxdxdu = -\sin x\,dx, v=exv = e^x

excosxdx=excosxex(sinx)dx\int e^x \cos x\,dx = e^x \cos x - \int e^x(-\sin x)\,dx

=excosx+exsinxdx= e^x \cos x + \int e^x \sin x\,dx


Substitute back:

Let I=exsinxdxI = \int e^x \sin x\,dx

I=exsinx(excosx+I)I = e^x \sin x - (e^x \cos x + I)

I=exsinxexcosxII = e^x \sin x - e^x \cos x - I

2I=ex(sinxcosx)2I = e^x(\sin x - \cos x)

I=ex(sinxcosx)2+CI = \frac{e^x(\sin x - \cos x)}{2} + C


When to Use Integration by Parts

Good Candidates

  1. Polynomial times exponential: xnexdx\int x^n e^x\,dx
  2. Polynomial times trig: xnsinxdx\int x^n \sin x\,dx or xncosxdx\int x^n \cos x\,dx
  3. Logarithm times polynomial: xnlnxdx\int x^n \ln x\,dx
  4. Inverse trig times polynomial: xarctanxdx\int x \arctan x\,dx
  5. Exponential times trig: exsinxdx\int e^x \sin x\,dx

⚠️ Common Mistakes

Mistake 1: Wrong Choice of uu

WRONG: For xexdx\int x e^x\,dx, choosing u=exu = e^x

This makes things harder! Differentiating exe^x doesn't simplify.

RIGHT: Choose u=xu = x (it simplifies when differentiated)


Mistake 2: Adding +C to vv

When finding v=dvv = \int dv, don't add +C yet!

The +C comes at the very end.


Mistake 3: Forgetting the Minus Sign

The formula is uvvduuv - \int v\,du (minus, not plus!)


Mistake 4: Not Simplifying

After applying the formula, you still need to evaluate vdu\int v\,du!


Mistake 5: Circular Without Solving

If you get the original integral back, don't give up!

Solve algebraically: I=II = \ldots - I2I=2I = \ldots


Integration by Parts vs U-Substitution

When to use what?

U-Substitution: Composite function with derivative present

  • 2x(x2+1)5dx\int 2x(x^2+1)^5\,dx
  • xex2dx\int xe^{x^2}\,dx

Integration by Parts: Product of different types of functions

  • xexdx\int x e^x\,dx (polynomial × exponential)
  • xsinxdx\int x \sin x\,dx (polynomial × trig)
  • lnxdx\int \ln x\,dx (logarithm)

Quick Reference

The Formula

udv=uvvdu\int u\,dv = uv - \int v\,du

LIATE Priority for uu

L - Logarithmic

I - Inverse trig

A - Algebraic

T - Trigonometric

E - Exponential

Steps

  1. Choose uu and dvdv (use LIATE)
  2. Find dudu and vv
  3. Apply formula
  4. Evaluate remaining integral
  5. Simplify and add +C

📝 Practice Strategy

  1. Use LIATE to choose uu (first in list) and dvdv (the rest)
  2. Find dudu by differentiating uu
  3. Find vv by integrating dvdv (no +C yet!)
  4. Write out the formula uvvduuv - \int v\,du before substituting
  5. Evaluate the new integral vdu\int v\,du
  6. Check: Does dudu simplify things? (If not, try different choice)
  7. Be ready to apply parts multiple times
  8. Watch for circular cases - solve algebraically
  9. Add +C at the very end
  10. Check by differentiating your answer

📚 Practice Problems

1Problem 1hard

Question:

Evaluate x2sinxdx\int x^2 \sin x\,dx using integration by parts.

💡 Show Solution

First application of integration by parts

Step 1: Choose uu and dvdv

Using LIATE: Algebraic before Trigonometric

u=x2u = x^2, dv=sinxdxdv = \sin x\,dx


Step 2: Find dudu and vv

du=2xdxdu = 2x\,dx

v=sinxdx=cosxv = \int \sin x\,dx = -\cos x


Step 3: Apply formula

x2sinxdx=uvvdu\int x^2 \sin x\,dx = uv - \int v\,du

=x2(cosx)(cosx)(2x)dx= x^2(-\cos x) - \int (-\cos x)(2x)\,dx

=x2cosx+2xcosxdx= -x^2 \cos x + 2\int x\cos x\,dx


Second application on xcosxdx\int x\cos x\,dx:

u=xu = x, dv=cosxdxdv = \cos x\,dx

du=dxdu = dx, v=sinxv = \sin x

xcosxdx=xsinxsinxdx\int x\cos x\,dx = x\sin x - \int \sin x\,dx

=xsinx(cosx)=xsinx+cosx= x\sin x - (-\cos x) = x\sin x + \cos x


Combine:

x2sinxdx=x2cosx+2(xsinx+cosx)+C\int x^2 \sin x\,dx = -x^2\cos x + 2(x\sin x + \cos x) + C

=x2cosx+2xsinx+2cosx+C= -x^2\cos x + 2x\sin x + 2\cos x + C

=(2x2)cosx+2xsinx+C= (2-x^2)\cos x + 2x\sin x + C

Answer: (2x2)cosx+2xsinx+C(2-x^2)\cos x + 2x\sin x + C

2Problem 2medium

Question:

Evaluate xexdx\int x e^x \, dx using integration by parts.

💡 Show Solution

Solution:

Integration by parts formula: udv=uvvdu\int u \, dv = uv - \int v \, du

Let u=xu = x and dv=exdxdv = e^x \, dx

Then: du=dxdu = dx and v=exv = e^x

Apply formula:

xexdx=xexexdx\int x e^x \, dx = x e^x - \int e^x \, dx

=xexex+C= x e^x - e^x + C

=ex(x1)+C= e^x(x - 1) + C

3Problem 3medium

Question:

Evaluate xlnxdx\int x\ln x\,dx.

💡 Show Solution

Step 1: Choose uu and dvdv

Using LIATE: Logarithmic before Algebraic

u=lnxu = \ln x (logarithmic - comes first in LIATE)

dv=xdxdv = x\,dx (algebraic)


Step 2: Find dudu and vv

du=1xdxdu = \frac{1}{x}\,dx

v=xdx=x22v = \int x\,dx = \frac{x^2}{2}


Step 3: Apply formula

xlnxdx=uvvdu\int x\ln x\,dx = uv - \int v\,du

=(lnx)(x22)x221xdx= (\ln x)\left(\frac{x^2}{2}\right) - \int \frac{x^2}{2} \cdot \frac{1}{x}\,dx

=x2lnx2x2dx= \frac{x^2\ln x}{2} - \int \frac{x}{2}\,dx

=x2lnx212xdx= \frac{x^2\ln x}{2} - \frac{1}{2}\int x\,dx

=x2lnx212x22+C= \frac{x^2\ln x}{2} - \frac{1}{2} \cdot \frac{x^2}{2} + C

=x2lnx2x24+C= \frac{x^2\ln x}{2} - \frac{x^2}{4} + C


Factor:

=x24(2lnx1)+C= \frac{x^2}{4}(2\ln x - 1) + C

Answer: x2lnx2x24+C\frac{x^2\ln x}{2} - \frac{x^2}{4} + C or x24(2lnx1)+C\frac{x^2}{4}(2\ln x - 1) + C

4Problem 4hard

Question:

Evaluate lnxdx\int \ln x \, dx.

💡 Show Solution

Solution:

Rewrite as (lnx)(1)dx\int (\ln x)(1) \, dx and use integration by parts.

Let u=lnxu = \ln x and dv=dxdv = dx

Then: du=1xdxdu = \frac{1}{x} dx and v=xv = x

lnxdx=xlnxx1xdx\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx

=xlnx1dx= x \ln x - \int 1 \, dx

=xlnxx+C= x \ln x - x + C

=x(lnx1)+C= x(\ln x - 1) + C

5Problem 5expert

Question:

Evaluate excosxdx\int e^x \cos x\,dx.

💡 Show Solution

This is a circular case!

First application:

u=cosxu = \cos x, dv=exdxdv = e^x\,dx

du=sinxdxdu = -\sin x\,dx, v=exv = e^x

excosxdx=excosxex(sinx)dx\int e^x \cos x\,dx = e^x\cos x - \int e^x(-\sin x)\,dx

=excosx+exsinxdx= e^x\cos x + \int e^x\sin x\,dx


Second application on exsinxdx\int e^x\sin x\,dx:

u=sinxu = \sin x, dv=exdxdv = e^x\,dx

du=cosxdxdu = \cos x\,dx, v=exv = e^x

exsinxdx=exsinxexcosxdx\int e^x\sin x\,dx = e^x\sin x - \int e^x\cos x\,dx


Substitute back:

Let I=excosxdxI = \int e^x\cos x\,dx

I=excosx+(exsinxI)I = e^x\cos x + (e^x\sin x - I)

I=excosx+exsinxII = e^x\cos x + e^x\sin x - I

2I=ex(cosx+sinx)2I = e^x(\cos x + \sin x)

I=ex(cosx+sinx)2+CI = \frac{e^x(\cos x + \sin x)}{2} + C


Answer: ex(cosx+sinx)2+C\frac{e^x(\cos x + \sin x)}{2} + C