3.23M

Category: $mathematics$ mathematics

Techniques of integration

1. TECHNIQUES OF INTEGRATION

Trigonometric Substitution
In this section, we will learn about:
The various types of trigonometric substitutions.

2. TRIGONOMETRIC SUBSTITUTION

Notice the difference between the substitution u = a2 – x2 and the
substitution x = a sin θ.
In the first, the new variable is a function of
the old one.
In the second, the old variable is a function of
the new one.
In general, we can make a substitution of the form x = g(t)
by using the Substitution Rule in reverse.
To make our calculations simpler, we assume g
has an inverse function, that is, g is one-to-one.

5. INVERSE SUBSTITUTION

Here, if we replace u by x and x by t in the Substitution Rule
we obtain:
This kind of
substitution
is
called
inverse
substitution.
We can make the inverse
substitution x = a sin θ, provided
that it defines
a one-to-one
function.
This can be accomplished
by restricting θ to lie in the
interval
[-π/2, π/2].

6. TABLE OF TRIGONOMETRIC SUBSTITUTIONS

Here, we list trigonometric substitutions that are effective for the
given radical expressions because of the specified trigonometric
identities.

7. TABLE OF TRIGONOMETRIC SUBSTITUTIONS

In each case, the restriction on θ is imposed to ensure that the
function that defines the substitution is one-to-one.
These are the same intervals used to defining
the inverse functions.

8. TRIGONOMETRIC SUBSTITUTION

Example 1

9.

Thus, the Inverse Substitution Rule gives:
Example 1
As this is an
indefinite integral,
we must return to
the original
variable x. This
can be done in
either of two ways.
One, we can use
trigonometric identities to
express cot θ in terms
of
sin θ = x/3.
Two, we can draw a diagram, where
θ is interpreted as an angle of a right
triangle.

10. TRIGONOMETRIC SUBSTITUTION

Example 1
Since sin θ = x/3, we label the opposite side and the hypotenuse as
having lengths x and 3.
Although θ > 0 here,
this expression for
cot θ is valid even
when θ < 0.
As sin θ = x/3, we have θ = sin-1(x/3).
Hence,

11. TRIGONOMETRIC SUBSTITUTION

Solution:
or
Example 2

12.

Solution
As the ellipse is symmetric with respect to both axes, the total area A
is four times the area in the first quadrant.

13.

Solution
To evaluate this integral, we substitute
x = a sin θ.
Then,
dx = a cos θ dθ.
To change the limits of integration,
we note that:
When x = 0, sin θ = 0;
so θ = 0
When x = a,
so θ = π/2
sin θ = 1;
Also, since 0 ≤ θ ≤ π/2,

14.

Therefore,
We have shown that the area of
an ellipse with semi-axes a and b
is
A=πab.
In particular, taking a =
b = r, we have proved
the famous formula
that the area of a circle
with radius r is πr2.

15.

Example 3
Solution:
Thus, we have:

16.

To evaluate this trigonometric integral, we put everything in terms of sin
θ and cos θ:
Therefore, making the
substitution u = sin θ, we
have:

17.

We use the figure to determine that:
Hence,

18.

Example 4
Solution
It
would be possible to use the
trigonometric substitution x = 2 tan θ;
However, the direct substitution u = x2 + 4
is simpler. Then, du = 2x dx
Note
Example 4 illustrates the fact that, even
when trigonometric substitutions are
possible, they may not give the easiest
solution.
You should look for a
simpler method first.