MAT137Y1 Midterm: 2006 Test 3 solution

• School: UTSG
• Department: Mathematics
• Course: MAT137Y1
• Professor: N/ A

Document Summary

3 sec3 x +c (by a simple substitution u = secx). Integrating by parts, let u = x, dv = 3x dx. Then dx = sec tan d , thus we have sec tan sec2 1 sec4 d = tan sec3 tan d = To integrate cos3 we take out one power of cos and apply a substitution: Substituting back, we can use triangles to see that if sec = x (1 sin2 )cos d = Z dx x2 1 x4 x2 1 x cos sin2 cos d = sin 1. 1, then sin = ( sin3 +c. Mat 137y 2006-2007 winter session, solutions to term test 3: evaluate the following integrals. Z (10%) (iii) sec3 xtanx dx. sec3 xtanxdx = x 3x dx. Z (10%) (iv) dx x3(x + 1)2 . Obtaining a common denominator on the right side (and ignoring the denominator) gives.