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

[tahayassen]

If we see the form $\sqrt { { a }^{ 2 }-{ x }^{ 2 } }$, we always set $x=asinθ$

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

 

[rock.freak667]

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

 

[tiny-tim]

hi tahayassen!

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

 

[Mark44]

If we see the form $\sqrt { { a }^{ 2 }-{ x }^{ 2 } }$, we always set $x=asinθ$

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).

 

[Sayajin]

If we see the form $\sqrt { { a }^{ 2 }-{ x }^{ 2 } }$, we always set $x=asinθ$

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 $y^2+x^2=r^2$. Your expression $\sqrt { { a }^{ 2 }-{ x }^{ 2 } }$ 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.

$x=asinz; dx=acoszdz$
$\int \sqrt{a^2-a^2sin^2z}acoszdz=$$a\int \sqrt{1-sin^2z}acoszdz$$= a^2\int \sqrt{cos^2z}coszdz =$$a^2\int cos^z dz$
$cos^2z=\frac{cos2z+1}{2}$$\frac{a^2}{2}\int cos2zdz+\frac{a^2}{2}\int dz$$= \frac{a^2sin2z}{4} + \frac{a^2z}{2} + C$

 

[cantRemember]

This contains most methods involving trigonometric substitution, and probably is the most usefull calculus reference sheet I have ever found. Link.

 

[HallsofIvy]

If we see the form $\sqrt { { a }^{ 2 }-{ x }^{ 2 } }$, we always set $x=asinθ$

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

We get this one from the identity $sin^2(t)+ cos^2(t)= 1$ so that $cos^2(t)= 1- sin^2(t)$, $cos(t)= \sqrt{1- sin^2(t)}$ so that the square root on the right does not occur on the left.

 

[dextercioby]

If we see the form $\sqrt { { a }^{ 2 }-{ x }^{ 2 } }$, we always set $x=asinθ$

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

I'd try $x = a\tanh t$