More U-Substitution:The “Double-U”Substitution with ArcTan(u)

Chapter 5.5

February 13, 2007

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 , ...

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∫∫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)

∫

Evaluate: sec2 (2t) tan2 (2t)dt∫

1

2tan2 (u)sec2 (u)du∫

1

2tan2 (2t)2sec2 (2t)dt∫

1

2sec2 (2t) tan2 (2t)2dt∫

u =2tdu=2dt

u =tan(2t)

du=2sec2 (2t)dt

w =tan(u)

dw=sec2 (u)du1

2w2dw∫

1

2u2du∫

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

9 + t2∫

Let :      u =t3

            du=13dt

=1

9

dt

1+t

3⎛⎝⎜⎞⎠⎟

2∫

=1

3

du

1 + u( )2∫ =

1

3tan−1 u( ) +C =

1

3tan−1 t

3⎛⎝⎜⎞⎠⎟

+C

We have the formula:dt

1+ t2∫ =tan−1 t( ) +C

Factor out the 9 in the expression 9 + t2:

dt

9 + t2∫ =dt

9 1+t2

9⎛⎝⎜

⎞⎠⎟

∫

=3

9

1

3dt

1+t

3⎛⎝⎜⎞⎠⎟

2∫

In general:

dt

a2 + t2∫

Let :      u =ta

            du=1adt

=1

a2

dt

1+t

a⎛⎝⎜

⎞⎠⎟

2∫

=1

a

du

1 + u( )2∫ =

1

atan−1 u( ) +C =

1

atan−1 t

a⎛⎝⎜

⎞⎠⎟

+C

We now have the formula:dt

a2 + t2∫ =1atan−1 t

a⎛⎝⎜

⎞⎠⎟+C

Factor out the a2 in the expression a2 + t2:

dt

a2 + t2∫ =dt

a2 1+t2

a2

⎛⎝⎜

⎞⎠⎟

∫

=a

a2

1

adt

1 +t

a⎛⎝⎜

⎞⎠⎟

2∫

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) + 25 = (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 ∫

Try these:

dx

x2 −x+ 2∫

dx

x2 + 7x−5 ∫

1+ x1+ x2∫  dx

In groups of two/three, use u-substitution to complete:

1.    −4z−5( )∫32 +

17z+ 4

dz

2. 7e5−2rdr∫

3.x

x32 +1( )

2

+12

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

∫ dx

4.z

z2 −4⎛⎝⎜

⎞⎠⎟∫ dz