Trigonometric Inverse Substitution: How do we know which substitutions to use?

In summary, Trigonometric inverse substitution is used when dealing with integrals containing expressions in the form of ax² + bx + c, with a non-zero constant and a perfect square. The most commonly used substitutions are sin θ = √(1 – cos²θ), cos θ = √(1 – sin²θ), and tan θ = √(1 + tan²θ). There are specific rules to follow when choosing a substitution, and the purpose of using trigonometric inverse substitution is to simplify integrals involving trigonometric functions. We can verify its correctness by differentiating the result and comparing it to the original integral.
  • #1
tahayassen
270
1
If we see the form [itex]\sqrt { { a }^{ 2 }-{ x }^{ 2 } }[/itex], we always set [itex]x=asinθ[/itex]

How do we know that it will work in advance? Just trial & error?
 
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  • #2
Work in advance to do what exactly? If you had to integrate that radical, then we would just try a substitution to eliminate the radical.

Putting x=asint or x=acost will both capitalize on the fact that sin2t+cos2t=1
 
  • #3
hi tahayassen! :smile:
tahayassen said:
How do we know that it will work in advance?

we don't know, but if a substitution will work, it'll be that one :wink:
 
  • #4
tahayassen said:
If we see the form [itex]\sqrt { { a }^{ 2 }-{ x }^{ 2 } }[/itex], we always set [itex]x=asinθ[/itex]

How do we know that it will work in advance? Just trial & error?
The context here is integration using trig substitution. It's similar to ordinary substitution except that you are using a trig function.

Trig substitutions are used most often for the square roots of a sum of squares or a difference of squares. For ##\sqrt { a^2 - x^2 } ##, the idea is that you are using a right triangle with an acute angle θ. The hypotenuse is of length a, and the side opposite the acute angle is of length x. From these relationships, we get
sin θ = x/a, or x = a sin θ.
Taking differentials, we get dx = a cos θ dθ.

If the quantity under the radical is x2 - a2, label the hypotenuse as x and one of the other sides as a.

If the quantity under the radical is a2 + x2, label the legs of the triangle as a and x, and the hypotenuse as ##\sqrt{a^2 + x^2}##. With this arrangement, tan θ = x/a (assuming the side across from the angle is labelled x).
 
  • #5
tahayassen said:
If we see the form [itex]\sqrt { { a }^{ 2 }-{ x }^{ 2 } }[/itex], we always set [itex]x=asinθ[/itex]

How do we know that it will work in advance? Just trial & error?

Its very intuitive actualy. The equation of a circle with radius r centered in the coordinate system is [itex] y^2+x^2=r^2[/itex]. Your expression [itex]\sqrt { { a }^{ 2 }-{ x }^{ 2 } }[/itex] is just a semi circle with radius a. In trigonometry circle with radius one is used to represent trigonometric functions using the coordinates. Since the radius is a and your coordinates are forming a right triangle with hypotenuse a you can use trig identities and substitute with sine of an angle multiplied by the radius of the circle.

The point in this is when you factor out a that 1-sin^2(z)=cos^2(z) .You will be left with a square of a single function in the square root. By using trigonometric identities for square of trig functions you can actualy simplify by finding the integral of trig function of the double angle.

[itex]x=asinz; dx=acoszdz[/itex]
[itex]\int \sqrt{a^2-a^2sin^2z}acoszdz=[/itex][itex]a\int \sqrt{1-sin^2z}acoszdz[/itex][itex]= a^2\int \sqrt{cos^2z}coszdz =[/itex][itex]a^2\int cos^z dz[/itex]
[itex]cos^2z=\frac{cos2z+1}{2}[/itex][itex]\frac{a^2}{2}\int cos2zdz+\frac{a^2}{2}\int dz[/itex][itex] = \frac{a^2sin2z}{4} + \frac{a^2z}{2} + C[/itex]
 
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  • #6
This contains most methods involving trigonometric substitution, and probably is the most usefull calculus reference sheet I have ever found. Link.
 
  • #7
tahayassen said:
If we see the form [itex]\sqrt { { a }^{ 2 }-{ x }^{ 2 } }[/itex], we always set [itex]x=asinθ[/itex]

How do we know that it will work in advance? Just trial & error?
We get this one from the identity [itex]sin^2(t)+ cos^2(t)= 1[/itex] so that [itex]cos^2(t)= 1- sin^2(t)[/itex], [itex]cos(t)= \sqrt{1- sin^2(t)}[/itex] so that the square root on the right does not occur on the left.
 
  • #8
tahayassen said:
If we see the form [itex]\sqrt { { a }^{ 2 }-{ x }^{ 2 } }[/itex], we always set [itex]x=asinθ[/itex]

How do we know that it will work in advance? Just trial & error?

I'd try [itex] x = a\tanh t [/itex]
 

Related to Trigonometric Inverse Substitution: How do we know which substitutions to use?

1. How do we determine when to use trigonometric inverse substitution?

Trigonometric inverse substitution is typically used when we encounter integrals that contain expressions in the form of ax² + bx + c, where a is a non-zero constant and c is a perfect square.

2. What are the common trigonometric substitutions used in integrals?

The most frequently used trigonometric substitutions are sin θ = √(1 – cos²θ), cos θ = √(1 – sin²θ), and tan θ = √(1 + tan²θ).

3. Can we use any trigonometric substitution, or are there specific rules to follow?

There are specific rules to follow when choosing a trigonometric substitution. We must identify a suitable substitution by observing the terms in the integral and choosing one that would simplify the expression.

4. What is the purpose of using trigonometric inverse substitution?

Trigonometric inverse substitution is used to simplify integrals that involve expressions with trigonometric functions. It allows us to transform an integral into a more manageable form that can be easily evaluated.

5. How do we verify the correctness of a trigonometric inverse substitution?

We can verify the correctness of a trigonometric inverse substitution by differentiating the result and comparing it to the original integral. If the derivative is equal to the integrand, then the substitution was successful.

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