Integration by Triangle Substitutions The Area of a Circle So far we have used the technique of u-substitution (ie, reversing the chain rule) and integration by parts (reversing the product rule) to etend the list" of functions that we can antidifferentiate Remember that we are in the antidifferentiation business because the Fundamental Theorem (FTC) says that if F is an antiderivative of f, then the area under f on the interval [a, b] is b a f () d = F(b) F(a) This theorem solves" the area problem, at least for those functions whose antiderivatives we know But even with our new integration techniques there are many integrals we cannot yet do, such as: EXAMPLE 7 Find the area of the semi-circle of radius r Since middle school we have been told that the area of a circle is πr, so the area of a semi-circle is πr But why is this formula valid? Recall that the entire circle of radius r centered at the origin is the set of points that satisfies + y = r It follows that the upper semi-circle is given by the function y = f () = r on the interval [ r, r] So the area we seek is r r r d It s scandalous but true: none of the techniques we ve developed so far will help us find an appropriate antiderivative Still r should remind us of right triangles, and we ve ecised one such triangle from the figure below f () = r r r The sides of the triangle are related to each other via the trigonometric functions of the angle We can use these to set up a fancy u-substitution, or more appropriately, a -substitution r = sin r r = r sin r = cos = r cos d = r sin d r r Figure 7: Left: A semi-circle with equation y = r Right: A corresponding right triangle Notice that each part of the original integrand can now be written in terms of the angle Of course, we need to write the -limits of integration as -limits When = r = r cos cos = = π = r = r cos cos = = This makes sense: changes from π to as you move around the semi-circle clockwise
math, techniques of integration iv triangle substitutions from r to r Now the integral can be rewritten in terms of and solved r r r d = π r sin ( r sin ) d = r sin d = r π π cos() d ( ) = r sin() [ π ( = r π )] ( ) = πr From this it follows that the area of a circle of radius r is πr 7 General Triangle Substitutions In general, triangle substitutions are based on, well, triangles! Right triangles, in particular Three different substitutions arise depending on the integrand and how the sides of the triangle are labeled The triangles are based on the corresponding forms of the square roots that arise from the Pythagorean theorem in these triangles The three cases are: + a, a, and a Case : Integrals involving + a In this case, a + must correspond to the hypotenuse of the right triangle (Why?) So with the legs of the triangle labeled as below, we have: + a a = a tan EXAMPLE 7 Calculate the following indefinite integral: d = a sec d + a = a sec + d Using the triangle substitution in the bo above with a = or a =, we have: + d = tan sec sec d = sec tan d = cos sin cos d cos = sin d = (sin ) cos d = (sin ) + c We must convert our answer back to a function of Look back at the triangle above Notice that sin = + So ( ) (sin ) + c = = + + c + One question you may have is how should you select which leg of the triangle should correspond to and which to a? The answer is that the substitution will work no matter which you use, but it will be easier with as the opposite side to angle If we had let be the adjacent side, then = a cot, a less familiar trig function EXAMPLE 7 Calculate the indefinite integral: 6 + d Trianglete Version: Mitchell-5/7/::
math, techniques of integration iv triangle substitutions Using the triangle substitution above with a = 6 or a =, we have: d = 6 + sec sec d = Case : Integrals involving a sec d = ln sec + tan + c = ln 6+ + + c In this case, must correspond to the hypotenuse of the right triangle (Why?) Again we have our choice of how to label the legs, one side a and the other a With the selection made below, = a sec What would equal if we had let a be the side opposite? = a sec a a d = a sec tan d a = a tan EXAMPLE 7 Evaluate the definite integral: d Use the triangle substitution above with a = or a = We also have to change limits = = sec = = = sec cos = = π/ Now proceed with the substitution: π/ tan π/ π/ d = sec sec tan d = tan d = sec d = tan π/ = π/ Case : Integrals involving a This was situation in the semi-circle eample In this case, a must correspond to the hypotenuse of the right triangle (Why?) Again we have our choice of how to label the legs, one side a and the other With the selection made below, = a sin In the circle eample we chose to label the legs in the other way because of the geometry involved Obviously it worked out, but it required us to carry along a minus sign throughout the problem The choice below is usually simpler a a = a sin d = a cos d a = a cos EXAMPLE 7 Calculate the following indefinite integral: 5 d Use the substitution above with a = 5 or a = 5 We use a reduction formula (see the Appendi to these notes) to do the integral d = 5 5 sin 5 cos 5 cos d = 5 sin d = 5 cos() d = 5 5 sin() + c Trianglete Version: Mitchell-5/7/::
math, techniques of integration iv triangle substitutions To solve for, look back at the original triangle where 5 = sin arcsin 5 = and 5 5 = cos The simplest way to finish the problem is to make use of a Double Angle Formula: sin() = sin cos Then 5 5 sin() + c = 5 5 ( sin cos ) + c = 5 arcsin 5 5 5 5 5 + c = 5 arcsin 5 5 + c 7 Additional Eamples In this last section we etend the triangle substitution idea to integrals that at first don t appear to have anything to do with triangles because a square root does not immediately appear in them EXAMPLE 7 Calculate the following indefinite integral: ( d + 6) / Well there is sort of a square root lurking in the background here In this case, ( + 6) / may be thought of as ( + 6) So the appropriate triangle is: + 6 6 = 6 tan d = 6 sec d + 6 = 6 sec Notice that one of the sides of the triangle is 6; we may not always have perfect squares Using the triangle substitution in the bo we have: 6 sec ( + 6) / d = 6 sec ( 6 sec ) d = 6 6 sec d = 6 EXAMPLE 7 Determine the following indefinite integral: sec d = 6 d This time we have to think of as ( ) to use a triangle cos d = 6 sin + c = 6 + 6 + c = sin d = cos d = cos Now we can rewrite the integral d = 8 cos ( cos ) cos d = cos d = = sec d ln sec() + tan d = ln + + c = ln + + c = ln (+) + c = ln + + c = ln + ln + c Trianglete Version: Mitchell-5/7/::
math, techniques of integration iv triangle substitutions 5 We will see this same integral again, shortly, and it will be solved very differently EXAMPLE 7 Evaluate the definite integral: / d This time the triangle is obvious, but care is required to label the sides correctly = sec d = sec tan d = tan You could do the indefinite integral and convert back to to avoid changing the limits But let s actually change them = = sec = sec = = = sec = sec cos = = π/ Now proceed with the substitution (and use the guidelines for integrating powers of sec ): / d = π/ ( sec ) tan sec tan d = π/ sec d 6 = π/ sec sec d 6 = π/ ( + tan ) sec d 6 u du = π/ {}}{{}}{ sec + tan sec d 6 = [ ] tan + tan π/ 6 ] [ = ( ) + 6 = 8 webwork: Click to try Problems through 75 Problems Try these problems A variety of techniques are required, not just triangle substitutions (a) 9 d (b) ( + ) / d (c) (d) d (e) d (f ) 9 (g) + d (h) + d (i) 5 + (j) 5 d (k) d (l) 5 + (m) d (n) d 5 5 + (o) 5 5 d (p) 5 d 6 d d d 5 d Find the arc length of the parabola f () = on the interval [, ] You will have to use a trig substitution Make sure that you switch the limits of the integration Trianglete Version: Mitchell-5/7/::
math, techniques of integration iv triangle substitutions 6 Answers Caution, it s easy to have made a typo in these answers Remember: +c ( ) (a) 9 arctan 9 (b) + (c) arcsin(/) (d) π ( (e) ) arctan 9 (f ) ( ) / (g) arctan(/) (h) ln + ( (i) ln + (j) ) 5 + 75 (k) 5( ) (m) 5 5 + (5 ) / (l) arcsin(/5) 5 + (n) ln + 5 5 (o) 5π/ Answer: +ln + ) (p) 5 5 76 Appendi: Common Trigonometric Formulas and Antiderivatives Below are listed several integral formulas for various powers of trig functions Degree Sine and Cosine Functions One simple way to do these is to use trig identities Know these (a) cos u du = + cos u du (b) sin u du = cos u du Low Powers of the Tangent and Secant Functions These are done with simple identities Know these sin u (a) tan u du = du = ln sec u + c cos u (b) tan u du = sec u du = tan u u + c (c) sec u du = sec u + sec u tan u sec u + tan u du = ln sec u + tan u + c Useful Double Angle Formula: sin() = sin cos Reduction Formulas for Large Powers These are verified using integration by parts Repeated application may be necessary (a) cos n u du = n cosn u sin u + n n cos n u du (b) sin n u du = n sinn u cos u + n n sin n u du (c) tan n u du = n tann u tan n u du (d) sec n u du = n secn u tan u + n n sec n u du 5 Degree Sine and Cosine Functions Again If we apply the reduction formulas as for the sine and cosine functions when n =, we get a different form of the earlier answer These new forms are better to use with indefinite integrals involving triangle substitutions because it is easier to convert back from u to the original variable (Know either these or those in #) (a) cos u du = cos u sin u + du = cos u sin u + u + c (b) sin u du = sin u cos u + du = sin u cos u + u + c Trianglete Version: Mitchell-5/7/::