Calculus II
Lesson 13: Integration Techniques Summary and Review
You can use the student package in Maple to practice your integration techniques. First load the student
package by typing
> with(student):
Then read over the help screens on changevar, intparts, and value, paying particular attention to the
examples at the bottom of the screens. Here are some examples.
A Substitution Problem:
Integration by substitution
f(g(x))
d
dx
du =
d
dx
g(x)
g(x)
is based on the chain rule. Thus if we have an integral which looks like
dx, then by make the change of variable g(x) = u, and letting
dx we have a new, perhaps simpler integral
f(u) du, to work on.
changevar
In Maple this is accomplished using the word
from the student package.
Find an antiderivative of
> F := Int(1/sqrt(1+sqrt(x)),x);
1
F :=
dx
1+
x
Let's try the change of variable sqrt(x) = u .
> G := changevar(sqrt(x)=u,F);
G :=
2u
du
1+u
This does not seem to help. Lets try 1 + sqrt(x)= u
> G := changevar(1+sqrt(x)=u,F);
G :=
2
1-2u+u
u
Now we can do it by inspection, so just finish it off.
> G := value(G);
2
du
G :=
4
2
u (-3 + u)
(-1 + u)
3 (-1 + u)
Now substitute back and add in the constant.
> F := subs(u=sqrt(x),G) + C;
(1/ 4)
F :=
4x
2
(-3 +
x)
(-1 +
3 (-1 +
x)
+C
x)
Integration by substitution is the method use try after you decide you can't find the antiderivative by
inspection.
An Integration by Parts Problem:
Integration by parts
u dv = uv -
is based on the product rule for derivatives. It is usually written
v du. It turns one integration problem into one which 'may' be more doable. Once you
decide to use parts, the problem is what part of the integrand to let be u.
Integrate
> F := Int(x^2*arctan(x),x);
F :=
2
x arctan(x) dx
2
The word is intparts . Let's try letting u = x .
> G := intparts(F,x^2);
2
G := x
2
1
x arctan(x) - ln(1 + x )
2
2x
-
x arctan(x) -
1
2
That was a bad choice. Try letting u = arctan(x)
> G := intparts(F,arctan(x));
3
1
x
3
G := arctan(x) x 3
2
dx
3 (1 + x )
This is much more promising. Split off the integral on the end.
> H := op(2,G);
3
H := -
x
2
2
ln(1 + x )
dx
3 (1 + x )
Now do a partial fractions decomposition of the integrand of H, using parfrac .
dx
> H:= Int(convert(integrand(H),parfrac,x),x);
1
H :=
x
x-
dx
2
3
3+3x
Now we can do it by inspection.
> H1 := 1/6*x^2 - 1/3*1/2*ln(1+x^2);
2
1 2 1
H1 := x - ln(1 + x )
6
6
Let's check this with the student value.
> simplify(value(H-H1));
1
- ln(3)
6
Note the difference of a constant, which is fine for antiderivatives.
ETAIL : The problem of choosing which part of the integrand to assign to u can often be solved quickly
by following the etail convention. If your integrand has an Exponential factor, choose that for u,
otherwise if it has a Trigonometric factor, let that be u, otherwise choose an Algebraic factor for u,
otherwise chose an Inverse trig function, and as a last resort choose u to be a logarithmic factor. Let dv
be what's left over.
A Trig Substitution:
Find an antiderivative of
> F := Int(x^3/sqrt(x^2+1),x);
3
x
F :=
dx
2
1+x
2
The presence of x + 1 suggests letting x = tan (t).
> G := changevar(x=tan(t),F,t);
G :=
2
tan (t)
3
2
1 + tan (t) dt
2
Now use the trig identity 1 + tan (t) = sec(t) .
> G := subs(sqrt(1+tan(t)^2)=sec(t),G);
G :=
3
tan (t) sec(t) dt
Another substitution into the integrand.
> G := subs(tan(t)^3 = (sec(t)^2-1)*tan(t),G);
2
G :=
(sec(t) - 1) tan (t) sec(t) dt
Let's make a change of variable,
> H := changevar(sec(t)=u,G);
2
H :=
u - 1 du
From here, we can do it by inspection.
> H := value(H);
1 3
H := u - u
3
Now unwind the substitutions.
> G := subs(u=sec(t),H);
G :=
1
3
sec(t) - sec(t)
3
> F := subs(t = arctan(x),G);
3
1
F := sec(arctan(x)) - sec(arctan(x))
3
> F := subs(sec(arctan(x))=sqrt(1+x^2),F) + C;
(3/ 2)
1
2
F := (1 + x )
3
-
2
1+x +C
Checking this calculation:
> F1 := int(x^3/sqrt(x^2+1),x);
F1 :=
1
2
x
2
1+x -
3
2
3
It looks different, but is it?
> simplify(F-F1);
C
Yes, but only by a constant.
A Partial Fractions Problem
Integrate the rational function
> y :=(4*x^2+x -1 )/(x^2*(x-1)*(x^2+1));
2
1+x
2
4x +x-1
y :=
2
2
x (x - 1) (1 + x )
First get the partial fractions decomposition
of y.
> y := convert(y,parfrac,x);
1
y :=
+
2
2
-
2
x-1
x
3+2x
1+x
We can almost do this by inspection, except for the last term.
> F := Int(y,x);
1
F :=
+
2
2
x-1
x
-
3+2x
dx
2
1+x
> F := expand(F);
F :=
1
2
x
dx + 2
1
1
dx - 3
dx - 2
2
x-1
1+x
x
1+x
Now we can do each one by inspection. So we'll just use value .
> F := value(F) + C;
2
1
F := - + 2 ln(x - 1) - 3 arctan(x) - ln(1 + x ) + C
x
Exercises:
Exercise: Use the student package to perform the following integrations.
cos(x)
dx
1 + sin(x)
3x-7
2
dx
(x - 1) (x - 2)
3
2
x sin(a x) dx
ln(x +
x) dx
dx
2
z
5
dz
2
z +1
1
dx
(3 x)
e
-1
x arcsin(2 x) dx
Exercise: Find the area of the region enclosed by the x-axis and the curve y = x sin(x) on the interval
[0, ]. Sketch the region. Then find the vertical line x = a that divides the region in half and plot it.
2
Exercise: Find the length of the graph of the parabola y = x from O(0,0) to P(10,100). Find the point
2
Q(a, a ) on the graph which is 10 units from O along the graph. Make a sketch, showing the points O, P,
and Q on the graph.
Exercise: Find the volume of the solid of revolution obtained by revolving the region trapped between
x
the the graph of y = e sin(x)on [0, n ]and the x-axis about the x-axis. Sketch a graph. Does this volume
approach a finite limit as n gets large?