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Unformatted text preview: Last Week: Two Techniques of Integration Last week we talked about two tools for integrating difficult functions: I Integration by Substitution (Change of Variable) Z x = b x = a f ( u ( x )) dx = Z u = u ( b ) u = u ( a ) f ( u ) dx du du = Z u = u ( b ) u = u ( a ) f ( u ) x ( u ) du = Z u = u ( b ) u = u ( a ) f ( u ) u ( x ) du I Integration by Parts Z udv = uv Z vdu Z b a udv = uv b a Z b a vdu Remember the following “tips” we talked about: I If you see a compound function, think about trying a substitution. Eg., Z 2 1 e 3 x dx = Z 6 3 e u 1 3 du = 1 3 Z 6 3 e u du I If you’re integrating the product of a polynomial times a function that antidifferentiates nicely (like e x or sin( x ) or cos( x )), try integration by parts. Eg., Z xe x dx = xe x Z e x dx = xe x e x + C I If you have a polynomial times ln( x ), ignore the last tip— differentiate the ln( x ) and antidifferentiate the polynomial. Eg., Z x 2 ln( x ) = x 3 3 ln( x ) Z x 3 3 1 x dx = 1 3 x 3 ln( x ) 1 3 Z x 2 dx More Tips We saw some instances where it was not obvious which tool(s) to use. I Trigonometric Identities: Eg., I Z tan( x ) dx = Z sin( x ) cos( x ) dx I Z sin 2 ( x ) dx = Z 1 2 cos(2 x ) 2 dx I Z sin 3 ( x ) dx = Z (1 cos 2 ( x )) sin( x ) dx which lead to a natural substitution. More Tips I The “ · 1” Trick , eg. Z tan 1 ( x ) dx = Z tan 1 ( x ) · 1 dx = x tan 1 ( x ) Z x · 1 1 + x 2 dx by integration by parts, and the integral R x 1+ x 2 dx on the right can be solved with the substitution u = 1 + x 2 . I Cyclic Integration by Parts, eg. Z sin( x ) cos( x ) dx = sin 2 ( x ) Z cos( x ) sin( x ) dx 2 Z sin( x ) cos( x ) dx = sin 2 ( x ) + C Z sin( x ) cos( x ) dx = 1 2 sin 2 ( x ) + C More Techniques Let’s consider three similar looking— but very different!— integrals: I Z 1 2 x 1 + x 2 dx = Z 2 1 du u = ln(2) Can be handled with the substitution u = 1 + x 2 , since du = 2 xdx is waiting for us in the numerator. I Z 1 2 1 + x 2 dx = 2 tan 1 ( x ) 1 = π 2 Cannot be solved with the substitution u = 1 + x 2 , but there is a “trigonometric substitution” we can do. I Z 1 2 1 x 2 dx = ∞ (diverges) Substitutions will not help you here! (At least, not directly.) This integral requires the method of “partial fractions”. Trigonometric Substitutions The name of the game is the following trigonometric identities: I 1 sin 2 ( θ ) = cos 2 ( θ ) I 1 + tan 2 ( θ ) = cos 2 ( θ ) cos 2 ( θ ) + sin 2 ( θ ) cos 2 ( θ ) = sec 2 ( θ ) I sec 2 ( θ ) 1 = tan 2 ( θ ) Notice that in each case, the right hand side of the identity is a perfect square. Trigonometric Substitution— R 1 1+ x 2 dx Remember the identity 1 + tan 2 ( θ ) = sec 2 ( θ ) Maybe we could use the substitution x = tan( θ ) to turn the denominator 1 + x 2 into sec 2 ( θ )? This will turn out to be useful because dx = sec 2 ( θ ) d θ . Watch: Z 1 1 + x 2 dx = Z 1 1 + tan 2 ( θ ) · sec 2 ( θ ) d θ = Z 1 sec 2 ( θ ) sec 2 ( θ ) d θ = Z d θ = θ + C = tan 1 ( x ) + C Trigonometric Substitution—...
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This note was uploaded on 02/10/2010 for the course MAT 132 taught by Professor Poole during the Spring '08 term at SUNY Stony Brook.
 Spring '08
 POOLE
 Calculus, Integration By Substitution

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