Www.le.ac.uk Integration by Substitution and by Parts Department of Mathematics University of Leicester

www.le.ac.uk

Integration by Substitution and by Parts

Department of MathematicsUniversity of Leicester

Content

Integration by Parts

Integration by Substitution

Another useful result

Introduction

Introduction

• Each function has its own integral (what it integrates to).

• We differentiated complex functions using the Product, Substitution and Chain Rules.

• With Integration, we will use ‘Integration by Parts’ and ‘Integration by Substitution’ to integrate more complex functions, using the simple functions as a starting point.

Next




Introduction

Integration by Substitution:Introduction

• Suppose we had to integrate something like:

• We know how to integrate , so what we really want to do is turn into a new variable, .

• Then we’ll have , which we know how to integrate.....

Next

dxx )12sin(

xsin12 x

u

usin




Introduction


• But this isn’t the end of the story.

• We’ve now got

• This doesn’t make any sense because we’ve got and inside the integral. We need to change everything so we’ve only got .

• How can we rewrite in terms of and ?

Next

udxsin

dx

xuu

u du




Introduction


• Well, , so

• Think of as a fraction, so we can

rearrange to get .

• Put this back into the integral:

Next

12 xu 2dx

du

dx

du

2

dudx

cuududuu cos

2

1sin

2

1

2sin




Introduction


• Then all you have to do is write everything

in terms of again:

Next

12 xu

x

cxcu )12cos(2

1 cos

2

1 So




Introduction

Integration by Substitution:General Method

Next

1. Choose a part of the function to substitute as .(choose a part that’s inside a function)

2. Find and rewrite in terms of .

3. Rewrite everything inside the integral in terms of .

4. Do the integration.

5. Rewrite in terms of .

dx

dudx du

u

u

x




Introduction

1. let (because it’s inside the √-function.)

2. , so

3.

Next

32 xu

xdx

du2

x

dudx

2

duux

duxxdxxx

2

1

233 22

Integration by Substitution:Example:

dxxx 32Integrate




Introduction

Integration by Substitution:Example

4.

5.

Next

cu

cu

duuduu

32

1

2

1

2

1 2

3

23

2

3

2

1

cx

cu

3

)3(

3

2

322

3

,32 xu




Introduction

Choose the appropriate substitutions:

Next

dxxx 2)1(

dxx

x 32

u

u




Introduction

Check Answers

Clear Hints and Answers

Show Answers

Hint

Hint

Integration by Substitution:Working with Limits

If there are limits on the integral, then you have 2 options:

1. Ignore the limits, work out the integral, then put them back in at the very end.

2. Rewrite the limits in terms of u, then just work with u and don’t change back to x at the end.

Next




Introduction


Consider the previous example:

1. Ignore the limits: We get as before.

Then put the limits in to get:

Next

cx

3

)3( 2

32

3

1

3

62

3




Introduction

3

2

2 3dxxx


2. Change the limits (using )

upper limit:

lower limit:

Then just forget and integrate for .

Next

32 xu633 so ,3 2 ux




Introduction

3

2

2 3dxxx

132 so ,2 2 ux

x u

Integration by Substitution:Question

Next

Calculate _____105

dxxx 222




Introduction

Check Answers

Clear Answers

Show Answers

Hint

Integration by Parts:Introduction

• Suppose we had to integrate something like:

• We know how to integrate and , but not when they are multiplied together.

• We have to use the following rule to integrate a product...

Next

dxxx sin

xsinx




Introduction

Integration by Parts:The formula

• are the two functions that

form the product.

• is what you get if you integrate .

Next

dx

dvxu and )(

)(xvdx

dv




Introduction

dxdx

duvuvdx

dx

dvu

Integration by Parts:The method

1. Decide which function you want to differentiate (this will be ) and which you want to integrate (this will be )

2. Find and from .

3. Put these into the formula.

Next

dx

dvxu and )(v

dx

dvu

dx

du




Introduction

dxdx

duvuvdx

dx

dvu

Next

dxxx sin

dxxx ln2

xdx

dvxu

xdx

dvxu

,sin

sin,

2

2

,ln

ln,

xdx

dvxu

xdx

dvxu

Integration by Parts:Choose the best way to split these integrals into parts...




Introduction

dxdx

duvuvdx

dx

dvu

Check Answers Clear Answers Show AnswersHint

Integration by Parts:Example

1. Differentiate because this will give a nice simple constant.

2.

3. So the integral is:

xx evdx

due

dx

dvxu 22

2

1and1Then.and

cexe

dxeexxx

xx 42 1

2

1

2

1 2222

Next

x

dxxe x2Integrate




Introduction

dxdx

duvuvdx

dx

dvu

Next

dxxx sin Integrate

dxxx ln Integrate 2





Introduction

dxdx

duvuvdx

dx

dvu

Reveal Answer

Reveal Answer

Hide Answers

Show Method

Show Method

Integration by Parts:Proof of formula

• To prove the formula, we start with the product rule for differentiation:

• Integrate both sides:

• Rearrange:

dxuvdxvuuv ''

'')( uvvuuvdx

dv

dxvuuvdxuv ''

Next




Introduction

dxdx

duvuvdx

dx

dvu

Integration by Parts:Another example

1. It doesn’t seem to matter which function we differentiate and which we integrate.

2. Let , so

3. Then the integral is:

Next dxxeex xx cossin

xdxe x sinIntegrate

xedx

dvxu ,sin xevx

dx

du ,cos




Introduction

dxdx

duvuvdx

dx

dvu


So

1. Choose the same way round as last time (otherwise we’ll reverse what we’ve just done).

2. Let , so

3. Then we getNext

dxxeexxdxe xxx cossinsin

xedx

dvxu ,cos xevx

dx

du ,sin

dxxexexe xxx sincossin

We perform Integration by Parts again on this term:




Introduction

dxdx

duvuvdx

dx

dvu

vu and


• So:

• ie.

• ie.

• If you can’t see what to do straight away, you should just play around with an integral until you spot something that works. Next

dxxexexexdxe xxxx sincossinsin

xxx xexexdxe cossinsin2

2

cossinsin

xxx xexe

xdxe




Introduction

dxdx

duvuvdx

dx

dvu


• We prove it using integration by substitution:

• Let . We get , so

• So the integral becomes:

cxfdxxf

xf

)(ln

)(

)(

Next

cxfcuduuxf

du

xf

xf

))(ln(ln 1

)()(

)(

)(xfu )(xfdx

du )(xf

dudx




Introduction

Decide which statements are true:

Next

7

2

35

22

32

2

33

3

3

5

32 )2()1(

)ln(sin sin

cos

sin2sin cos

3

)1()1(

)(

)( form theof is

2

1

3

dxxx

cxdxx

x

dxxxxxdxxx

cx

dxx

dxxf

xfdx

x

x

ce

dxexx




Introduction

(red is false, green is true)

Check Answers

Clear Answers

Show Answers

Conclusion

• We can integrate simple functions using the rules for those functions.

• We can integrate more complex functions by substituting for part of the function, or by integrating a product.

Next




Introduction

Documents

Www.le.ac.uk Integration by Substitution and by Parts Department of Mathematics University of Leicester