Trigonometric Substitution

E.G. Find the area of a circle of radius 2 units.

solution : this is easy, simply find since is one half (the top half) of a unit circle (see below!). The only trouble is: what the heck is an antiderivative of ? Not so easy!

> plot(sqrt(4-x^2),x=-2..2,title=`sqrt(4-x^2)`,thickness=2);

It would be easy if we could make the under the radical sign into a perfect square since we could then simply take the square root and be done with the nasty square root sign. This would happen if became because then we would get which would turn into .

> I1:=2*Int(sqrt(4-x^2),x=-2..2);with(student):

> I2:=changevar(x=2*sin(theta),I1,theta);

> value(I2);

Here's T.M.G.A. of :

> int(sqrt(4-x^2),x);

Trigonometric Substitutions

The substitution used above is an example of trig substitution . It is not the same as the earlier type of substitution or change of variable technique. Then the substituion was of the form where as opposed to trig substitution which is of the form , (and depends on the function f being one-to-one so it has an inverse so that . For this reason, trig substitution could be called inverse substitution )

The trigonometric Pythagorean Identity, in its three equivalent foms (all of which involve perfect squares ) are what motivate the three trigonometric substitutions listed in the table below.

In the following three examples you will see each form. I'll give you the trigonometrically transformed integral and Maple's version of an antiderivative of the original integrand.

first form:

> I1:=Int(x^3*sqrt(9-x^2),x);I2:=changevar(x=3*sin(theta),I1,theta);

> value(I1);

second form:

> I1:=Int(1/(x*sqrt(x^2+3)),x);I2:=changevar(x=sqrt(3)*tan(theta),I1,theta);value(I1);

Now, the Maple answer involves something unfamiliar called the inverse hyperbolic tangent function. The answer I get by hand is

> Me:=-sqrt(3)/3*ln(abs(sqrt((x^2+3))/x+sqrt(3)/x));Maple:=value(I1);

> simplify((Me-Maple));evalf(subs(x=1.3,Me-Maple));

third form: find the area under y = above .

This is not quite one of the forms in the table. If you observe though that we would like to become (why??) you will see that the appropriate substitution is obtained from letting i.e. .

> plot(sqrt(9*x^2-4)/x,x=0..2);

> I1:=Int(sqrt(9*x^2-4)/x,x=2/3..2*sqrt(2)/3);I2:=changevar(x=2/3*sec(theta),I1,theta);

> value(I1);evalf(%);

Two that look different.

1.

> I1:=Int(1/sqrt(9*x^2+6*x-8),x);

This one looks impossible until you complete the square under the radical sign by realizing that = . We then get a manageable form by letting See the two big steps below (you fill in the blanks):

> I2:=simplify(changevar(w=x+1/3,I1,w));

> I3:=changevar(w=sec(theta),I2,theta);

> value(I1);

2.

> I1:=Int(exp(1)^t*sqrt(9-exp(1)^(2*t)),t);

> I2:=changevar(u=exp(1)^t,I1,u);

> I3:=simplify(%);

> I4:=value(I3);

> answer1:=subs(u=exp(1)^t,I4);

answer1 above is what you get by hand . If all you know is Maple then you will get the answer below:

> Int(exp(1)^t*sqrt(9-exp(1)^(2*t)),t)=int(exp(1)^t*sqrt(9-exp(1)^(2*t)),t);;

#37 page 489

Find the area bounded by and .

> bottom:=2-x;top:=x^2*sqrt(4-x^2);

> plot({top,bottom},x=-2..2,colour=[black,magenta],thickness=2);

> top-bottom;

> intercepts:=evalf(solve(top-bottom));

> area:=Int(top-bottom,x=intercepts[4]..intercepts[3]);

> value(area);