13.2
13.3
13.4
13.5
Use the change of variable
Use the change of variable
A triangle is shown in Figure CD-13.2 for a change of variable
A triangle is shown in Figure CD-13.2 for a change of variable
We are beginning to wallow a little too deeply in trig.
The point of the previous example could simply be: change the limits of integration when you change variable and differential.
However, it is possible to change back to u, and the previous exercise gives you a start on the trig skills needed to do this.
Problem CD-13.1
A Constant
Is
13.7
13.8
13.5.1
Combining these facts we have,
and
Sketch the graph
for
. What geometrical shape is shown in your graph? What is the area of one fourth of a circle of unit radius?
. (Check your symbolic computation geometrically).
with an appropriate differential to show that
How large can we take v?
with an appropriate differential to show that
How large can we take v?
Suppose we make the change of variable
(as in the integration above). Express
in terms of u by using Figure CD-13.1 and TOA.
. Express
and
in terms of v by using the figure and the Pythagorean Theorem.
. Express
and
in terms of w by using the figure and the Pythagorean Theorem.
= CAH and
= TOA
Use the change of variable
with an appropriate differential to show that
Also, use the change of variable
with an appropriate differential to show that
? Ask the computer to Plot
and
. Why do they look alike? How do they differ? Do the graphs of
and
look alike? How do they differ?
, so, when the cosine is positive,
and
, so
from 0 to
, with equal areas above and below the
-axis.
when we let
. There is an important algebraic detail when we write
and chose
to put
in this range.
It is also true that
for this range of
, so that
can first be reduced to a multiple of
by taking u=x/a and writing
. This means that the expression
with the substitutions u=(x/a)2 and
(provided cosine is positive on the interval).
converts
into
, provided sine is positive.
Also note that sine is positive over a different range of angles than cosine.
, then u2+1 becomes
and
provided secant is positive.
and
. If we really want the antiderivative of
, then we must express all this in terms of u.
must first be written in terms of functions of
, (recall the addition formula for sine), 
idea.
From SOH-CAH-TOA, if we take a right triangle with hypotenuse 1 and opposite side u, then
is the adjacent side of this triangle.
Using the Pythagorean Theorem, we have
and
