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

  1. Jan 24, 2013 #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?
  2. jcsd
  3. Jan 24, 2013 #2


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    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
  4. Jan 24, 2013 #3


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    hi tahayassen! :smile:
    we don't know, but if a substitution will work, it'll be that one :wink:
  5. Jan 24, 2013 #4


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    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).
  6. Jan 24, 2013 #5
    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]\frac{a^2}{2}\int cos2zdz+\frac{a^2}{2}\int dz[/itex][itex] = \frac{a^2sin2z}{4} + \frac{a^2z}{2} + C[/itex]
    Last edited: Jan 24, 2013
  7. Jan 25, 2013 #6
    This contains most methods involving trigonometric substitution, and probably is the most usefull calculus reference sheet I have ever found. Link.
  8. Jan 25, 2013 #7


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    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.
  9. Jan 26, 2013 #8


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    I'd try [itex] x = a\tanh t [/itex]
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