q3solns - Mathematics 104, Quiz 3, Friday, Feb 29, 2008 1....

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Unformatted text preview: Mathematics 104, Quiz 3, Friday, Feb 29, 2008 1. Find sec3 tan3 d. sec3 tan3 d = sec2 tan2 (sec tan ) d = sec2 (sec2 - 1)(sec tan ) d Making the substitution u = sec we get u2 (u2 - 1) du = . . . = sin3 d = cos6 sec5 sec3 - +C 5 3 Alternatively you can rewrite the integral in terms of sin and cos : 1 - cos2 sin d cos6 u2 - 1 du. u6 and then make the substitution u = cos to get 2. Find dx . + 2x + 10)5/2 (x2 Complete the square to replace x2 + 2x + 10 by (x + 1)2 + 9 and then make the substitution du u = x + 1 to get . Then u = 3 tan gives 2 + 9)5/2 (u sin3 +C 3 u so Returning to the original variable, if u = 3 tan and x + 1 = u then sin = 2+9 u we get 1 (x + 1)3 x+1 - +C 34 x2 + 2x + 10 ( x2 + 2x + 10)3 3 sec2 1 d = 4 2 )5/2 3 (9 sec cos3 d = 1 34 (1 - sin2 ) cos d = 1 34 sin - 3. Find x2 dx. 9 - x2 Make the substitution x = 3 sin to get 9 sin2 3 cos d = 9 3 cos 9 1 - cos 2 d = 2 2 - sin 2 2 +C 9 - x2 to get 3 Now we use the fact that sin 2 = 2 cos sin and the fact that cos = 9 2 arcsin(x/3) - 9 - x2 x 3 3 +C = 9 1 arcsin(x/3) - x 2 2 9 - x2 + C ...
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