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Lecture 4 &amp; 5

# Lecture 4 &amp; 5 - MATH 118 LECTURES 4& 5...

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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 let’s 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 )! Let’s 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 ∈ [- π/ 2 , π/ 2]. A fortunate consequence of this is that cos( u ) is never negative over u ∈ [- π/ 2 , π/ 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....
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Lecture 4 &amp; 5 - MATH 118 LECTURES 4& 5...

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