This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MATH 118, LECTURES 4 & 5: Trigonometric Substitutions 1 Trigonometric Substitution In this lecture, we are going to use some of our trigonometric identities to simplify integrals. The identities we use relate to the form of derivatives we know: the derivatives of inverse trigonometric functions. d dx arcsin( x ) = 1 1 x 2 1 x 2 1 sin 2 ( x ) = cos 2 ( x ) d dx arctan( x ) = 1 1 + x 2 1 + x 2 1 + tan 2 ( x ) = sec 2 ( x ) d dx arcsec( x ) = 1 x x 2 1 x 2 1 sec 2 ( x ) 1 = tan 2 ( x ) . Our approach will be a little different than in previous lectures. Rather than simply integrating the above expressions, we will try a different intuition. Consider the integral integraldisplay 1 1 x 2 dx. We recognize very quickly that we are looking at the integral of the derivative of arcsin( x ), so the fundamental theorem of calculus tells us that integraldisplay 1 1 x 2 dx = arcsin( x ) + C, but lets assume we are not quite so enlightened. How could we solve the integral directly? We might as well start by trying a substitution, but which substitution will work? The trouble point is the square root, since it does not distribute over the terms contained within it. We would much rather have a single term under the square root; however, it can be easily checked that the substitution u = 1 x 2 will not work. We appear to be stuck. There is however, another substition which is capable of simplifying 1 x 2 into a single term. The trick is to recall our trig identities from last week, in particular sin 2 ( x ) + cos 2 ( x ) = 1 (1) 1 1 + tan 2 ( x ) = sec 2 ( x ) . (2) If we rearrange identity (1) we get cos 2 ( x ) = 1 sin 2 ( x ), so if we choose x = sin( u ) (or u = arcsin( x )) then 1 x 2 becomes 1 sin 2 ( u ) which simplifies to cos 2 ( u )! Lets try this: integraldisplay 1 1 x 2 dx = integraldisplay 1 radicalbig 1 sin 2 ( u ) cos( u ) du = integraldisplay cos( u )  cos( u )  du = integraldisplay 1 du = u + C = arcsin( x ) + C. This is exactly what we expected! Note that I ignored the absolute value on the cos( x ) in the denominator when I cancelled the fraction. We are justified in doing this only because the transformation x = sin( u ) is defined over the range of its inverse, u = arcsin( x ), which is u [ / 2 , / 2]. A fortunate consequence of this is that cos( u ) is never negative over u [ / 2 , / 2], so that  cos( u )  = cos( u ) wherever the integral is defined. So we can evaluate integrals containing troublesome terms of the form 1 x 2 by substituting x = sin( u ). We can, in fact, handle more general cases by adjusting by constants to get the following principle....
View
Full
Document
This note was uploaded on 04/14/2009 for the course ENGINEERIN MATH 118 taught by Professor Soares during the Spring '09 term at Waterloo.
 Spring '09
 Soares

Click to edit the document details