General Contents
Detailed Contents
Index
Programmed tutorial: Integration of the arcsin: Example 1
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Find the integral
.
General Contents
Detailed Contents
Index
What is a good method to try on a complicated integrand like
arcsin
(
4
x
) where we know its derivative but not its integral?
Integration by Parts frequently works.
It requires doing another integral, but that second one may be simpler.
State integration by parts in general, using
u
and
dv
.
What should we take as
u
?
If we take
u
=
arcsin
(
4
x
), we will be able to use the derivative of the
arcsin
in forming
du
.
Using a general variable,
w
, state the derivative of
arcsin
(
w
).
Use this to find the
du
we need for
our Integration by Parts.
Note the parentheses ( ) in the radical.
Complete the numerator.
State what
dv
is equal to in our problem.
As usual, it is equal to the rest of the problem.
Here we have
dv
=
dx
Find
v
for our problem.
As always, we get
v
by integrating
dv
.
What do we get?
We get
Substitute these results into our integral.
becomes
Equation 1
Now we have the challenge of doing this new integral.
What method might work here?
Since there is an
x
2
in the denominator, and the combination
x*
dx
in the numerator, substitution may help.
Let’s work on just the second integral, including the negative sign:
Equation 2
We’ll substitute, using a variable
t
= 1 – (4
x
)
2
. Find
dt
.
We get
Express
x*
dx
in terms of
t
and
dt
.
Rewrite the integral in Equation 2 in terms of
t
and
dt
.
What rule of integration can we use here?
The Power Rule.
What is the power here?
The power is
Do the integral.
Express this partial result in terms of
x
using our definition above.
With
t
= 1 – (4
x
)
2
,
we get
Combine this result with our earlier results in Equation 1 to solve the original problem.
Equation 1 becomes
Equation 3
How can we check this result?
We can differentiate the result,
and compare with the integrand of the original problem. What rule can we use on the first term?
The Product Rule.
Do the differentiation.
We get
Simplify this result and use a radical in the third term.
Does it check?
Since the third term reduces to be the negative of the first term, this result is the same.
Let’s try solving this same problem by using the
Table of Integrals
.
What is the number of the integral in the table which applies?
Number 41 applies here.
What is the value of “
a
” for this problem?
a
= 4
What is the result of the integral using this value of “
a
”
Does this agree with our previous result in Equation 3?
Yes.
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