Notes PDF
Introductory Problems
1. Find ∫ex/[1 + ex] dx
Answer
1. ln (1 + ex) + c
Solution
1. Substitute u= ex. Then du= ex dx, and we get ∫ex/[1 + ex] dx = ∫1/(1 + u) du = ln(1 + u) + c = ln (1 + ex) + c
2. Find ∫cos5(x) dx
Answer
1. sin(x) – 2/3 sin3(x) + 1/5 sin5(x)
Solution
1. We can write: cos5(x) = [cos2(x)]2 cos(x) = [1 – sin2(x)]2 cos(x) Now we can substitute u = sin(x), so du = cos(x) dx, and we get ∫cos5(x) dx = ∫[1 – u2]2 du= ∫(1 – 2 u2 + u4) du= u – 2/3 u3 + 1/5 u5 = sin(x) – 2/3 sin3(x) + 1/5 sin5(x)
3.Find ∫cos2(x) dx
Answer
1. x/2 + 1/4 sin(2x) + c
Solution
1. Just as we did in the notes with sin2(x), we can use the formula for cos(2x) to help us out: cos(2x) = cos2(x) – sin2(x)= cos2(x) – [1 – cos2(x)] = 2 cos2(x) – 1 Now we turn it around: cos2(x) = 1/2[1 + cos(2x)] This we know how to integrate: ∫cos2(x) dx = ∫1/2[1 + cos(2x)] dx = x/2 + 1/4 sin(2x) + c
4. Use the formula ∫(1 – x2)-1/2 dx = arcsin(x) + c to find ∫(4 – x2)-1/2 dx
Answer
1. Arcsin(x/2) + c
Solution
1. Our first goal is to turn the “4” into a “1”, so we pull 4 our of the parentheses: (4 – x2)-1/2= [4 (1 – x2/4)]-1/2= 1/2 [1 – (x/2)2]-1/2 Now we substitute u = x/2, so du = dx / 2, and get ∫(4 – x2)-1/2 dx = ∫[1 – u2]-1/2 du= arcsin(u) + c = arcsin(x/2) + c
5. Use trig substitution to find ∫1/(1 + x2) dx
Answer
1. Arctan(x) + c
Solution
1. Use our trig substitution table, and substitute x = tan(u). As written in the notes: 1 + x2 = 1 + tan2(u) = 1/cos2(u) In exercises for Algebra of derivatives we calculated the derivative of tan(x) using the product rule: dx = 1/cos2(u) du The two go very well together: 1/(1 + x2) dx = cos2(u) dx = du Easy to integrate: ∫1/(1 + x2) dx = ∫du = u + c = arctan(x) + c