Download Integration Review : Simplify the Integrand if Possible.

Integration Review
Simplify the integrand if possible.
– p. 1/12
Integration Review
Simplify the integrand if possible.
Look for an obvious or a clever substitution.
– p. 1/12
Integration Review
Simplify the integrand if possible.
Look for an obvious or a clever substitution.
Decide how you are going to integrate:
* Integration by Parts
* Trigonometric Integration
* Trigonometric Substitution
* Partial Fractions.
– p. 1/12
Integration Review
Simplify the integrand if possible.
Look for an obvious or a clever substitution.
Decide how you are going to integrate:
* Integration by Parts
* Trigonometric Integration
* Trigonometric Substitution
* Partial Fractions.
If none of the above work, try to manipulate the
integrand to transform it into an easier form which you
can apply the above techniques.
– p. 1/12
Integration Review : Simplify the Integrand if Possible.
!
(x + sin x)2 dx
– p. 2/12
Integration Review : Simplify the Integrand if Possible.
!
(x + sin x)2 dx
! 2
= x + 2x sin x + sin2 xdx
– p. 2/12
Integration Review : Simplify the Integrand if Possible.
!
(x + sin x)2 dx
! 2
= x + 2x sin x + sin2 xdx
! 2
1 3
1. x dx = x + C
3
!
!
2. 2x sin xdx = 2 x sin xdx
!
= 2(−x cos x − (− cos x) · 1dx) by Integration by Parts,
= −2x
x + 2 sin x + C
! cos
3. sin2 xdx
! 1 − cos(2x)
1
1
=
dx = (x − sin(2x)) + C
2
2
2
– p. 2/12
Integration Review : Simplify the Integrand if Possible.
!
(x + sin x)2 dx
! 2
= x + 2x sin x + sin2 xdx
! 2
1 3
1. x dx = x + C
3
!
!
2. 2x sin xdx = 2 x sin xdx
!
= 2(−x cos x − (− cos x) · 1dx) by Integration by Parts,
= −2x
x + 2 sin x + C
! cos
3. sin2 xdx
! 1 − cos(2x)
1
1
=
dx = (x − sin(2x)) + C
2
2
2
Therefore,
!
1 3
1
1
2
(x+sin x) dx = x −2x cos x+2 sin x+ x− sin(2x)+C
3
2
4
– p. 2/12
Integration Review : Simplify the integrand if Possible.
! sec x cos(2x)
dx
Try
2 sin x + sec x
– p. 3/12
Integration Review: u-Substitution.
!
1
√ dx
√
x+x x
– p. 4/12
Integration Review: u-Substitution.
!
1
√ dx
√
x+x x
√
1
Let u = x so that du = √ dx and x = u2 .
2 x
– p. 4/12
Integration Review: u-Substitution.
!
1
√ dx
√
x+x x
√
1
Let u = x so that du = √ dx and x = u2 .
2 x
Then
!
!
1
1
√ dx =
√ dx
√
x+x x
(1 + x) x
– p. 4/12
Integration Review: u-Substitution.
!
1
√ dx
√
x+x x
√
1
Let u = x so that du = √ dx and x = u2 .
2 x
Then
!
!
1
1
√ dx =
√ dx
√
x+x x
(1 + x) x
!
!
1
1
=
2du = 2
du
2
2
1+u
1+u
– p. 4/12
Integration Review: u-Substitution.
!
1
√ dx
√
x+x x
√
1
Let u = x so that du = √ dx and x = u2 .
2 x
Then
!
!
1
1
√ dx =
√ dx
√
x+x x
(1 + x) x
!
!
1
1
=
2du = 2
du
2
2
1+u
1+u
√
−1
−1
= 2 tan u + C = 2 tan
x+C
– p. 4/12
Integration Review: u-Substitution.
Try
!
Try
!
√
−1
tan
xdx
e
√
x dx
– p. 5/12
Integration Review: u-Subsitution and Trigonometric Substitution.
!
1
√
dx
x2 4x2 + 1
– p. 6/12
Integration Review: u-Subsitution and Trigonometric Substitution.
!
1
√
dx
x2 4x2 + 1
√
Try with u = 4x2 + 1 and complete the integration with
a trigonometric substitution.
– p. 6/12
Clever u-Substitution
! ln(tan x)
dx
sin x cos x
– p. 7/12
Clever u-Substitution
! ln(tan x)
dx
sin x cos x
Let u = ln(tan x) so that
cos x
1
1
1
2
· sec xdx =
·
dx =
dx.
du =
2
tan x
sin x cos x
sin x cos x
– p. 7/12
Clever u-Substitution
! ln(tan x)
dx
sin x cos x
Let u = ln(tan x) so that
cos x
1
1
1
2
· sec xdx =
·
dx =
dx.
du =
2
tan x
sin x cos x
sin x cos x
Then
! ln(tan x)
!
dx = udu
sin x cos x
– p. 7/12
Clever u-Substitution
! ln(tan x)
dx
sin x cos x
Let u = ln(tan x) so that
cos x
1
1
1
2
· sec xdx =
·
dx =
dx.
du =
2
tan x
sin x cos x
sin x cos x
Then
! ln(tan x)
!
dx = udu
sin x cos x
1
1 2
= u + C = (ln(tan x))2 + C .
2
2
– p. 7/12
Clever u-Substitution.
Try
!
Try
!
cos x cos3 (sin x)dx
1
1 + 2ex
− e−x
dx
– p. 8/12
Integration by Parts
! ln(x + 1)
dx
2
x
– p. 9/12
Integration by Parts
! ln(x + 1)
dx
2
x
! 1 1
1
= − ln(x + 1) − (− )
dx by Integration by Parts,
x
x x+1
– p. 9/12
Integration by Parts
! ln(x + 1)
dx
2
x
! 1 1
1
= − ln(x + 1) − (− )
dx by Integration by Parts,
x
x x+1
!
1
1
= − ln(x + 1) +
dx
x
x(x + 1)
– p. 9/12
Integration by Parts
! ln(x + 1)
dx
2
x
! 1 1
1
= − ln(x + 1) − (− )
dx by Integration by Parts,
x
x x+1
!
1
1
= − ln(x + 1) +
dx
x
x(x + 1)
! 1
1
1
−
dx by Partial Fractions,
= − ln(x + 1) +
x
x x+1
– p. 9/12
Integration by Parts
! ln(x + 1)
dx
2
x
! 1 1
1
= − ln(x + 1) − (− )
dx by Integration by Parts,
x
x x+1
!
1
1
= − ln(x + 1) +
dx
x
x(x + 1)
! 1
1
1
−
dx by Partial Fractions,
= − ln(x + 1) +
x
x x+1
1
= − ln(x + 1) + ln |x| − ln |x + 1| + C
x
– p. 9/12
Integration by Parts
! ln(x + 1)
dx
2
x
! 1 1
1
= − ln(x + 1) − (− )
dx by Integration by Parts,
x
x x+1
!
1
1
= − ln(x + 1) +
dx
x
x(x + 1)
! 1
1
1
−
dx by Partial Fractions,
= − ln(x + 1) +
x
x x+1
1
= − ln(x + 1) + ln |x| − ln |x + 1| + C
x
x
= ln
+C
1
(x + 1) x (x + 1)
– p. 9/12
Trigonometric Integration
!
tan4 xdx
– p. 10/12
Trigonometric Integration
!
tan4 xdx
!
= (tan2 x)2 dx
– p. 10/12
Trigonometric Integration
!
tan4 xdx
!
= (tan2 x)2 dx
!
= (sec2 x − 1)2 dx
– p. 10/12
Trigonometric Integration
!
tan4 xdx
!
= (tan2 x)2 dx
!
= (sec2 x − 1)2 dx
!
= sec4 x − 2 sec2 x + 1dx
– p. 10/12
Trigonometric Integration
!
tan4 xdx
!
= (tan2 x)2 dx
!
= (sec2 x − 1)2 dx
!
= sec4 x − 2 sec2 x + 1dx
!
= (tan2 x + 1) sec2 x − 2 sec2 x + 1dx
– p. 10/12
Trigonometric Integration
!
tan4 xdx
!
= (tan2 x)2 dx
!
= (sec2 x − 1)2 dx
!
= sec4 x − 2 sec2 x + 1dx
!
= (tan2 x + 1) sec2 x − 2 sec2 x + 1dx
1
= tan3 x + tan x − 2 tan x + x + C
3
– p. 10/12
Trigonometric Integration
!
tan4 xdx
!
= (tan2 x)2 dx
!
= (sec2 x − 1)2 dx
!
= sec4 x − 2 sec2 x + 1dx
!
= (tan2 x + 1) sec2 x − 2 sec2 x + 1dx
1
= tan3 x + tan x − 2 tan x + x + C
3
1
= tan3 x − tan x + x + C .
3
– p. 10/12
Trigonometric Integration
Try
!π
4
0
tan5 θ sec3 θdθ
– p. 11/12
Partial Fractions
Try
!
1
dx
4
x(x + 1)
– p. 12/12