MATH 31B: MIDTERM 2 REVIEW. sin 2 x = 1 cos(2x) dx = x 2 sin(2x) 4. + C = x 2. dx = x sin(2x) + C = x sin x cos x

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1 MATH 3B: MIDTERM REVIEW JOE HUGHES. Evaluate sin x and cos x. Solution: Recall the identities cos x = + cos(x) Using these formulas gives cos(x) sin x =. Trigonometric Integrals = x sin(x) sin x = cos(x) + C = x sin x cos x and similarly + cos(x) cos x = = x + sin(x) + C = x sin x cos x + + C The last step of both integrals used the formula sin(x) = sin x cos x.. Evaluate sin x cos 3 x. Solution: The key idea for evaluating trig integrals with odd powers is to use the identity sin x + cos x = to try to set up a substitution. In this case, we get sin x cos 3 x = sin x cos x cos x = sin x( sin x) cos x Now the integral is written in a way that makes a u-substitution possible. Let u = sin x, so = cos x. Then the integral becomes u ( u ) = u u = u3 3 u5 5 + C = sin3 x sin5 x + C 3 5 One of the challenges of trig integrals is that the answer can take several different forms, depending on how many trig identities one chooses to use. So as a sanity check, let s take the derivative of our answer: + C d [ sin 3 x 3 sin5 x 5 + C ] = sin x cos x sin x cos x = sin x cos x sin x sin x cos x

2 JOE HUGHES = sin x cos x sin x( cos x) cos x = sin x cos x sin x cos x + sin x cos 3 x which is what we started with. 3. Evaluate cos x. = sin x cos 3 x Solution: To compute this integral, we can repeatedly use the identity from problem. Thus ( + cos(x) ) cos x = (cos x) [ ] = = + cos(x) + cos (x) = [ + cos(x) + + cos(x) ] and cos x = + cos(x) + + cos(x) = 3 + cos(x) + cos(x) 8 8 = 3 8 x + sin(x) + 3 sin(x) + C. Evaluate x 9 x.. Trig Substitutions Solution: Make the substitution x = 3 sin θ, so that = 3 cos θ dθ. Then x 9 sin = θ 3 cos θ dθ = 9 sin θ dθ 9 x 3 cos θ = 9 [θ sin θ cos θ] + C using the result from the previous section. Now θ = sin x 3, sin θ = x 3, and cos θ = 9 x 3, so x 9 x = 9. Evaluate x x. [ sin x 3 x 3 9 x 3 ] + C = 9 sin x 3 x 9 x + C Solution: At first glance, it doesn t look like we can do a trig substitution. But observe that (x 6) = x x + 36

3 hence MATH 3B: MIDTERM REVIEW 3 = x x x x = 36 (x 6) = 36 (x 6) 36 u by taking u = x 6. Now we can do a trig substitution: let u = 6 sin θ. Then = 6 cos θ dθ, hence 6 cos θ = 36 u 6 cos θ dθ = θ + C = u sin 6 + C = x 6 sin + C 6. Evaluate x (x ) (x 3). 3. Partial Fractions Solution: The numerator has degree less than the denominator, so we can use partial fractions. Write x (x ) (x 3) = A x + A (x ) + B x 3 After clearing denominators, we get x = A (x )(x 3) + A (x 3) + B(x ) Now plug in x = to obtain = A, so A =. Similarly, if we plug in x = 3 then we get = B. To determine A, let s look at the coefficient of x on both sides. On the left-hand side, this coefficient is zero, while on the right-hand side the coefficient is A + B. A = B =. So our integral becomes x (x ) = (x 3). Evaluate 5x (x+)(x +). x (x ) + = ln x 3 ln x + x + C x 3 Solution: When we do partial fractions with an irrecible quadratic factor like x +, the corresponding term in the expansion should have a linear polynomial in the numerator. So write 5x (x + )(x + ) = A x + + Bx + C x + Now clear the denominators to get 5x = A(x + ) + (Bx + C)(x + )

4 JOE HUGHES Next, plug in x = : or A =. 5( ) = A( + ) To find B, look at the coefficient of x. On the left-hand side this coefficient is zero, while on the right hand side it is A + B. B = A =. Finally, to find C we look at the constant coefficient on both sides to get = A + C, so C = A =. 5x (x + )(x = + ) 3. Evaluate sec θ tan θ dθ. x + x + + x = + x + + = ln x + + ln(x + ) + tan (x) + C x x + + x + Solution: One might be tempted to try to use the trig identity sec θ = + tan θ, but I don t think that helps here. So instead let s start with a substitution: let u = tan θ, so that = sec θ dθ. Then sec θ tan θ dθ = u = Now we can proceed using partial fractions. Write (u )(u + ) = A u + B u + and clear denominators to obtain = A(u + ) + B(u ) (u )(u + ) Plugging in u = gives = A, so A =. Similarly, plugging in u = gives B =. (u )(u + ) = u u + = ln u ln u + + C and plugging in u = tan θ gives sec θ tan θ dθ = ln tan θ ln tan θ + + C. Improper Integrals. Determine whether x 7 8 converges, and if so, evaluate it: Solution: The function x 7 8 has an infinite discontinuity at x =, so this is an improper integral. We can evaluate it using the limit definition: [ ] = lim = lim 8x 8 = lim 8 8a 8 = 8 a + a + a a + x 7 8 x 7 8 a

5 MATH 3B: MIDTERM REVIEW 5 the integral converges, and the value of the integral is 8.. Determine whether x + x converges. Solution: x x for x, hence x + x x and x +x x We can break our integral up into two pieces: x + x x + x = x = lim R x + x + x + x for x. The first integral is just some finite number, so it doesn t affect the convergence or divergence of the original integral. For the second integral, the comparison test gives [ ] R ln(x) ln(r) ln() = diverges, which shows that also diverges. x + x x + x = lim R 5. Numerical Integration. Find N such that cos x T N 5 Solution: The formula for the error bound for the trapezoid rule is given by where K = max x [a,b] f (x). Error(T N ) K (b a) 3 N If f(x) = cos x, then f (x) = sin x and f (x) = cos x. hence K = max cos x = x [, ] Error(T N ) ( ( ))3 N = N

6 6 JOE HUGHES We want the error to be at most 5. Thus and solving for N gives so any N 9 will do. N 5 N 5 9. How could we handle the last part without using a calculator? Using the fact that 5 5 = we see that any N = = suffices. 6. Arc length For t [, ], let f(t) be the arclength of the curve y = x t ( t) xt+ (t + ) on [, ]. Find the maximum of f(t) on [, ] and the point at which it is attained. Solution: Recall that the arclength formula is ( dy ) + By the power rule, hence + dy = x t xt ( dy ) x t = + + xt = x t + + xt ( x t = + xt ) Plugging this into the arc length formula gives f(t) = x t + x t = ( + t) + ( t) = t f is increasing on [, ], so has a maximum of 3 at t =.

7 MATH 3B: MIDTERM REVIEW 7 7. Fluid Force. Find the fluid force on a metal plate bounded by the parabola y = x and the line y =. The surface of the liquid is the line y =, and the liquid has pressure ρ. Solution: The formula for fluid force is F = ρg yf(y) dy where f(y) is the width at depth y. In this case, f(y) = y. F = ρg y y dy = ρg y 3 dy = ρg [ 5 y 5 8. Taylor Polynomials ] = 5 ρg. Let f(x) = e x. Find the third Taylor polynomial T 3 (x) for f centered at x =, and use the error bound to find the maximum value for f(.) T 3 (.). Solution: Recall that T 3 (x) = f() + f ()x + f () x + f (3) () x 3 6 we need to take 3 derivatives: Plugging in x =, we get f (x) = e x f (x) = e x f (3) (x) = e x Next, recall that the error bound formula is T 3 (x) = x + x x3 6 f(.) T 3 (.) K.! = K where K is an upper bound for f () (x) on [,.]. To get the best possible error bound, we take K = max f () (x) x [,.] Now f () (x) = e x, which is a decreasing function. hence K = f(.) T 3 (.) max f () (x) = e = x [,.].6 6 On the midterm, I think that it would suffice to leave the answer as.