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MATH 255 - Calculus II Trigonometric Substitution (Section 7.4)
Our objective: To “get rid of square roots” of sums (
a^2 + b^2 ) or differences (
c^2 − b^2 ) in integrands by replacing them with equivalent trigonometric functions. Hopefully, the resulting integrand will be one that is easily integrated.
Before we can start discussing how to go about using trigonometric substitution in integrals, we must first reacquaint ourselves with some key concepts in geometry and trigonometry:
Pythagorean’s Theorem We know, when dealing with right triangles that
(short side 1 )^2 + (short side 2 )^2 = (hypotenuse)^2
or, if we label our hypotenuse as c and our shorter sides as a and b, this becomes
a^2 + b^2 = c^2.
From this identity, we get two possible substitutions:
c =
a^2 + b^2 or a =
c^2 − b^2
as the figure below illustrates.
Figure 1: Rewriting Sides of the Right Triangle to Include Square Roots of Sums or Differences
Trigonometric Identities
sin(θ) = opphyp csc(θ) = hypopp
cos(θ) = (^) hypadj sec(θ) = hypadj
tan(θ) = oppadj cot(θ) = adjopp
Using Trigonometric Substitution Remember, we use trigonometric substitution to help us evaluate an integral with a square root of a sum or difference of perfect squares. Consider using trigonometric substitution to evaluate the indefinite integral: ∫ 1 x^2
4 − x^2
dx
Steps To Generally Follow A Specific Example
- Draw a right triangle.
- Identify the “problematic square root” term in your inte- grand.
Problem term:
4 − x^2
- If it has a sum of squares, let the square root be the hypotenuse (c =
a^2 + b^2 ).
- If it has a difference of squares, let the square root be one of the short sides (a =
c^2 − b^2 ).
Then use the terms under the square root to determine the missing values on the other sides.
- There should be a single x-term on one of the sides (give or take a constant). Find the simplest trigonometric function that incorporates the side with the x term in it, and the side with the next simplest expression.
x is on “opposite” side and hypotenuse of 2 is most simple so, sine relates these: sin(θ) =
x 2
- Solve the identity equation you just had for x. x = 2 sin(θ)
- Differentiate this equation so you have dx in terms of dθ. dx = 2 cos(θ)dθ
- If there are terms in the integral that we have not solved for yet, find a way to represent them in terms of our trigonomet- ric functions.
We can replace x’s with 2 sin(θ)’s. We need to represent
4 − x^2. Note cos(θ) =
√ 4 −x 2 2 so √ 4 − x^2 = 2 cos(θ)
- Make the substitutions back into the original integral. DON’T FORGET to sub in completely dx! ∫^ 1 x^2
4 − x^2
dx =
2 cos(θ) (2 sin(θ))^2 ∗ (2 cos(θ))
dθ
Simplify the expression as much as possible.
4 sin^2 (θ)
dθ =
csc^2 (θ)dθ
- Complete the integration process; you should end up with an answer in terms of θ’s. (^) =^1 4
(− cot(θ)) + C
- You need to convert your solution to x’s.
- If there are any θ’s outside of trig functions, solve your original equation relating x and trig functions for θ.
- If there are trig functions of θ in your solution, “read your solution off of the triangle,” using the identities relating sides of the triangle to trig functions.
There are no θ’s by themselves, just a cotangent. And cot(θ) = adjopp =
√ 4 −x^2 x , so ∫ dx x^2
4 − x^2
4 − x^2 x
+ C
