Fall09-Pfinal-sol

Fall09-Pfinal-sol - MAT126 Fall 2009 Practice Final The...

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Unformatted text preview: MAT126 Fall 2009 Practice Final The actual Final exam will consist of twelve problems 1 Problem 1 1. Evaluate integraltext π/ 2 π/ 3 sin 3 ( x ) cos 2 ( x ) dx 2. Evaluate integraltext sin 2 ( x ) dx Solution: 1. First, recognize that the integrand consists of sin( x ) raised to an odd power, multiplied by cos( x ) raised to an even power. Such a function is a prime candidate for the substitution u = cos( x ) (so that one of the sin( x ) can be used for the change of variable, and the rest may be expressed in terms of cos( x ) via the identity sin 2 ( x ) = 1 − cos 2 ( x )). Therefore we substitute u = cos( x ), whereby du = − sin( x ) dx , and we can write (noting that u = cos( π/ 3) = 1 / 2 when x = π/ 3, and u = cos( π/ 2) = 0 when x = π/ 2) integraldisplay x = π/ 2 x = π/ 3 sin 3 ( x ) cos 2 ( x ) dx = integraldisplay x = π/ 2 x = π/ 3 ( − sin 2 ( x ) cos 2 ( x ))( − sin( x ) dx ) = integraldisplay x = π/ 2 x = π/ 3 ( − 1 + cos 2 ( x ))(cos 2 ( x ))( − sin( x ) dx ) = integraldisplay u = u = 1 / 2 ( u 2 − 1)( u 2 ) du = integraldisplay u = u = 1 / 2 ( u 4 − u 2 ) du = u 5 5 − u 3 3 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle u = u = 1 / 2 = parenleftBigg (0) 5 5 − (0) 3 3 parenrightBigg − parenleftBigg (1 / 2) 5 5 − (1 / 2) 3 3 parenrightBigg = − 2 − 5 5 + 2 − 3 3 2. There are two ways to approach this problem. The first is to use the identity cos(2 x ) = cos 2 ( x ) − sin 2 ( x ) = 1 − 2 sin 2 ( x ) to get sin 2 ( x ) = 1 2 (1 − cos(2 x )) 2 This implies that integraldisplay sin 2 ( x ) dx = integraldisplay 1 2 (1 − cos(2 x )) dx = integraldisplay 1 2 dx − integraldisplay 1 2 cos(2 x ) dx = 1 2 x − 1 2 integraldisplay cos(2 x ) dx = 1 2 x − 1 2 sin(2 x ) 2 + C = 1 2 x − 1 4 sin(2 x ) + C The second approach is to use an integration by parts, setting u = sin x and dv = sin xdx ; this means that du = cos xdx and v = − cos x . Therefore, integration by parts gives integraldisplay sin 2 ( x ) dx = (sin x )( − cos x ) − integraldisplay ( − cos x )(cos xdx ) = − sin x cos x + integraldisplay cos 2 ( x ) dx Now we plug in the identity cos 2 ( x ) = 1 − sin 2 ( x ) to get integraldisplay sin 2 ( x ) dx = − sin x cos x + integraldisplay (1 − sin 2 ( x )) dx = − sin x cos x + integraldisplay dx − integraldisplay sin 2 ( x ) dx Adding integraltext sin 2 ( x ) dx to both sides of the equation gives 2 integraldisplay sin 2 ( x ) dx = − sin( x ) cos( x ) + integraldisplay dx = − sin( x ) cos( x ) + x + C so that integraldisplay sin 2 ( x ) dx = 1 2 x − 1 2 sin( x ) cos( x ) + C Notice that both approaches give the same answer, because of the identity sin(2 x ) = 2 sin( x ) cos( x )....
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This note was uploaded on 10/21/2010 for the course MAT 126 taught by Professor Sutherland during the Spring '07 term at SUNY Stony Brook.

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Fall09-Pfinal-sol - MAT126 Fall 2009 Practice Final The...

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