MATH 1120H Midterm: MATH1120H-Test-Practice-Solutions
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Mathematics 1110h calculus i: limits, derivatives, and integrals. [12 = 4 3 each: z tan2(x) cos3(x) dx, z ln(10) te t dt. 2 s ln ( s: z z2 + 1 z2 1, z 3. After some preliminary algebra, we will use the substitution u = sin(x), so du = cos(x) dx. cos(x)(cid:19)2. Z tan2(x) cos3(x) dx =z (cid:18) sin(x) cos3(x) dx =z sin2(x) cos2(x) cos3(x) dx. =z sin2(x) cos(x) dx =z u2 du = 3 sin3(x) + c (cid:3: this is obviously an improper integral: 8(cid:21) = 0 + z2 1(cid:19) dz = z 1 dz +z. 2 dz. dz = z (cid:18) z2 1 z2 1 dz = z z2 1 + 2 z2 1: z z2 + 1 z2 1. The rst integral is easy, to do the second we need to apply the method of partial fractions. z2 1 = (z 1)(z + 1), so, for some constants a and b, we have.
