more u-substitution february 17, 2009. substitution rule for indefinite integrals if u = g(x) is a...

More U-Substitution

February 17, 2009

Substitution Rule for Indefinite Integrals

If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then

f ' g(x)( ) g'(x)dx=∫ f u( )∫ du= f g x( )( ) +C

Substitution Rule for Definite Integrals

If g’(x) is continuous on [a,b] and f is continuous on the range of u = g(x), then

f g(x)( ) g'(x)dxa

b

∫ = f(u)dug(a)

g(b)

∫

Indefinite Integrals by Substitution

1) Choose u.

2) Calculate du.

3) Substitute u.Arrange to have du in your integral also.(All xs and dxs must be replaced!)

4) Solve the new integral.

5) Substitute back in to get x again.

€

du =du

dxdx

Techniques of Integration so far…

1. Use Graph & Area ( )

2. Use Basic Integral Formulas

3. Simplify if possible (multiply out, separate fractions…)

4. Use U-Substitution…..

r2 −x2 , x , ...

Choosing u Try to choose u to be an inside function.

(Think chain rule.) Try to choose u so that du is in the

problem, except for a constant multiple.

Choosing uFor

u = 3x + 2 was a good choice because

(1) 3x + 2 is inside the exponential.

(2) The derivative is 3, which is only a constant.

Doesn’t Fit AllWe can’t use u–substitution to solveeverything. For example:

Let u = x2

du = 2x dx

We need 2x this time, not just 2.

We CANNOT multiply by a variable to adjust our integral.

We cannot complete this problem

Doesn’t Fit AllFor the same reason, we can’t do the following by u–substitution:

But we already knew how to do this!

Doesn’t Fit AllFor the same reason, we can’t do the following by u–substitution:

But we already knew how to do this!

Morals No one technique works for everything. Don’t forget things we already know!

There are lots of integrals we will never learn how to solve…

That’s when Simpson’s Rule, the Trapezoidal Rule, Midpoint,….. need to be used to estimate…...

Definite Integrals

Evaluate: dx

x −2( )20

4

∫

Definite Integrals: Why change bounds?

Can simplify calculations. (no need to substitute back to the original variable)

Try:

Change of bounds: when x = -5, u = 25 - (-5)2=0 when x = 5, u = 25 - (5)2 = 0

x 25 −x2dx−5

5

∫ −1

225 − x2 (−2x)dx

−5

5

∫u =25 −x2

du=−2xdx

−1

2udu

0

0

∫

Definite Integrals & bounds

x 25 −x2dx−5

5

∫ −1

225 − x2 (−2x)dx

−5

5

∫

u =25 −x2

du=−2xdx

−1

2udu

0

0

∫

Examples

cos(3x)dx∫ 1

3cos( 3x ) 3dx∫

x 4 + x2( )dx∫

sin( x )

xdx∫

ln x( )( )2

xdx∫

1

24 + x2( ) 2xdx∫

2sin( x )

2 xdx∫

ln(x)( )2

xdx∫

1

3cos( u ) du∫

1

2u du∫

2 sin( u ) du∫

u( )2du∫

Compare the two Integrals:

x 5 −xdx1

3

∫5 −xdx1

3

∫  u =5−xdu=−dx

5 −xdx1

3

∫ x 5 −xdx1

3

∫

− 5 − x (−dx1

3

∫ )when x =1, u=5−1=4when x=3, u=5−3=2

udu2

4

∫

2u3

2

32

4

=2(4)

3

2

3−

2(2)3

2

3=

16

3−

4 2

3

Extra “x”

Notice that the extra ‘x’ is the same power as in the substitution:

 u =5−xdu=−dx

x 5 −xdx1

3

∫

− x 5 − x (−dx1

3

∫ )

when x =1, u=5−1=4when x=3, u=5−3=2 (5 −u) udu

2

4

∫

=10u

3

2

3−

2u5

2

52

4

=10(4)

3

2

3−

2(4)5

2

5

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

−10(2)

3

2

3−

2(2)5

2

5

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Extra “x”

 x =5−u

(5u1

2 −u32 )du

2

4

∫

Compare:

x 3+ x2( )4dx∫ x2 3+ x2( )

4dx∫ x3 3+ x2( )

4dx∫

Let :   u =3+ x2

du=2xdx

1

23+ x2( )

42xdx∫

1

2u( )4 du∫

1

2x 3+ x2( )

42xdx∫

1

2x2 3+ x2( )

42xdx∫

Since :       u =3+ x2

We have: x2 =u−3

1

2(u −3) u( )4 du∫

Still have an extra “x” that can’t be related to the substitution.

U-substitution cannot be used for this integral

Evaluate:

tdt

1+ 9t4∫

Let :      u =3t2

du=6tdt

1

6

6tdt

1+ 3t2( )2∫ =

1

6

du

1 + u( )2∫ =

1

6tan−1 u( ) + C

=1

6tan−1 3t 2( ) + CReturning to the original variable “t”:

Evaluate:

dt

1+ 2t−5( )2∫

Let :      u =2t−5 du=2dt

1

2

2dt

1+ 2t−5( )2∫ =1

2

du

1 + u( )2∫ =

1

2tan−1 u( ) + C

=1

2tan−1 2t − 5( ) + CReturning to the original variable “t”:

Evaluate:

dt

25 + 3t+1( )2∫

Let :      u =3t+1 du=3dt

1

3

3dt

52 + 3t+1( )2∫ =1

3

du

52 + u( )2∫ =

1

3⎛⎝⎜⎞⎠⎟

1

5tan−1 u

5⎛⎝⎜

⎞⎠⎟

+ C

=1

15tan−1 3t +1

5⎛⎝⎜

⎞⎠⎟

+ CReturning to the original variable “t”:

Use:

dt

25 + t−1( )2∫ =dt

t2 −2t+ 26∫

It’s necessary to know both forms:

t2 - 2t +26 and 25 + (t-1)2

t2 - 2t +26 = (t2 - 2t + 1) + (-1+26)

= (t-1)2 + 25

Completing the Square:

Comes from (a +b)2 =a2 + 2ab+b2

=x2 + x             +3

(2ab)    2xb =x⇒ b=12

a =x

(x +12)2 =x2 + 2x

12+

12

⎛⎝⎜

⎞⎠⎟

2

+1

4−

1

4

= x2 + x +1

4⎛⎝⎜

⎞⎠⎟

+ −1

4+

12

4⎛⎝⎜

⎞⎠⎟

=x2 + x +1

4

=(x +1

2)2 +

11

4

x2 + x+ 3

Use to solve:

How do you know WHEN to complete the square?

x2 + x+ 3=(x+12)2 +

114

dx

x2 + x+ 3∫

Ans: The equation x2 + x + 3 has NO REAL ROOTS(Check b2 - 4ac)

If the equation has real roots, it can be factored and later we will use Partial Fractions to integrate.

Evaluate:

dx

3x2 + x+1∫

dx

2x2 −7x+ 5 ∫

Even Powers of Sine

1

21−cos2x( ) dx∫

sin4 x dx∫

sin2 x dx∫

1

21−cos2x( )⎛

⎝⎜⎞⎠⎟

2

dx∫

Even Powers of Cosine

1

21+ cos2x( ) dx∫

cos4 x dx∫

cos2 x dx∫

1

21+ cos2x( )⎛

⎝⎜⎞⎠⎟

2

dx∫

Odd Powers use forms of:

sin3 x dx∫

cos2 x + sin2 x=1

sin2 x =1−cos2 x cos2 x=1−sin2 x

Save one sinx for the du

If then (adjust for -)

sin2 x sin x dx∫ =− 1 − cos2 x( )(−sin x) dx∫=− 1 − u2( )du∫

du=-sinxdx u =cosx

Replace the remaining even powers of sinx with sin2x = 1 - cos2x

Odd Powers use forms of:

cos3 x dx∫

cos2 x + sin2 x=1

sin2 x =1−cos2 x cos2 x=1−sin2 x

Save one cosx for the du

If then

cos2 x cos x dx∫ = 1 − sin2 x( )(cos x) dx∫= 1 − u2( )du∫

du=cosxdx u =sinx

Replace the remaining even powers of cosx with cos2x = 1 - sin2x