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Trigonometric substitution is a process in which the substitution of a trigonometric function into another expression takes place. It is used to evaluate integrals or it is a method for finding antiderivatives of functions that contain square roots of quadratic expressions or rational powers of the form (where p is an integer) of quadratic expressions. Examples of such expressions are:
or or etc.
The method of trigonometric substitution may be called upon when other more common and easier-to-use methods of integration have failed. Trigonometric substitution assumes that you are familiar with standard trigonometric identities, the use of differential notation, integration using u-substitution, and the integration of trigonometric functions.
x = f(θ)
⇒ dx = f'(θ)dθ
Read in Detail: Calculus in Maths
We use trigonometric substitution in the following cases,
Expression | Substitution |
|---|---|
a2 + x2 | x = a tan θ |
a2 - x2 | x = a sin θ |
x2 - a2 | x = a sec θ |
| x = a cos 2θ |
| x = α cos2θ + β sin2θ |
We can apply the trigonometric substitution method as discussed below,
Let's consider an example of the Integral involving a2 - x2.
Example:
Lets put, x = a sinθ
⇒ dx = a cosθ dθ
Thus, I =
⇒ I =
⇒ I =
⇒ I = θ + c
As, x = a sinθ
⇒ θ =
⇒ I =
Let's consider an example of the Integral involving x2 + a2.
Example: Find the integral
Solution:
Lets put x = a tanθ
⇒ dx = a sec2θ dθ, we get
Thus, I =
⇒ I =
⇒ I =
⇒ I = + c
As, x = a tanθ
⇒ θ =
⇒ I = + c
Let's consider an example of the Integral involving a2+ x2.
Example: Find the integral of
Solution:
Lets put, x = a tanθ
⇒ dx = a sec2θ dθ
Thus, I =
⇒ I =
⇒ I =
⇒ I =
⇒ I =
⇒ I =
⇒ I =
⇒ I =
⇒ I =
⇒ I =
⇒ I =
Let's consider an example of the Integral involving x2 - a2.
Example: Find the integral of
Let's put, x = a secθ
⇒ dx = a secθ tanθ dθ
Thus, I =
⇒ I =
⇒ I =
⇒ I =
⇒ I =
⇒ I =
⇒ I =
⇒ I =
⇒ I =
⇒ I =
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Problem 1: Find the integral of
Solution:
Taking 5 common in denominator,
⇒ I =
⇒ I =
According to theorem 1, a =
⇒ I = + c
⇒ I = + c
Problem 2: Find the integral of
Solution:
Taking √2 common in denominator,
⇒ I =
⇒ I =
According to theorem 1, a = 2
⇒ I = +c
⇒ I = +c
Problem 3: Find the integral of
Solution:
By rearranging, we get
Here taking, a = 3 and x = 3 sin θ
⇒ dx = 3 cos θ dθSubstituting these values,
I =
⇒ I =
⇒ I =
⇒ I = 243
⇒ I = 243Lets take,
u = cos θ
⇒ du = -sin θ dθSubstituting these values, we get
⇒ I = 243
⇒ I = -243
⇒ I = -243
⇒ I = -243As, u = cos θ and x = 3 sinθ
⇒ cos θ =
⇒ u =
⇒ u =Hence,I = -243
⇒ I = -243 + c
Problem 4: Find the integral of
Solution:
Taking 9 common in denominator,
I =
⇒ I =According to theorem 2, a =
⇒ I =
⇒ I =
Problem 5: Find the integral of
Solution:
Taking 4 common in denominator,
I =
⇒ I =According to theorem 3, a =
⇒ I =
⇒ I =
⇒ I =
⇒ I =
Problem 6: Find the integral of .
Solution:
Taking 2 common in denominator,
I =
⇒ I =According to theorem 4, a =
I =
⇒ I =
⇒ I =
⇒ I =
⇒ I =
Problem 7: Find the integral of .
Solution:
After rearranging, we get
I =
⇒ I =
⇒ I =
⇒ I =According to the theorem 2, we have
x = x-and a =
⇒ I =
⇒ I =
1. Evaluate:
2. Find:
3. Compute:
4. Evaluate:
5. Find:
6. Compute:
7. Evaluate:
8. Find:
9. Compute:
10. Evaluate: