Math Calculus II Homework # Due Date Solutions

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Alfred Doyle
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1 Math 35 - Calculus II Homework # Due Date Solutions Part : Problems from sections 7.3 and 7.4. Section 7.3: 9. + d We will use the substitution cot(θ, d csc (θ. This gives + + cot (θ d csc (θdθ. cot(θ Net, some algebraic manipulations: + cot (θ csc (θdθ cot(θ tan(θ csc 3 (θdθ csc (θ csc(θ tan(θdθ ( + cot (θ csc(θ tan(θdθ csc(θ tan(θ + csc(θ cot(θdθ sec(θdθ + csc(θ cot(θdθ. Both of the final integrals above we can do. So we get sec(θdθ + csc(θ cot(θdθ ln (sec(θ + tan(θ + csc(θ + D. We now have to draw a triangle. Since cot(θ, we get that tan(θ (we could do that without the triangle, and sec(θ +, csc(θ +. So we now have ( + + d ln D.

2 7. d ( + + First, we complete the square: gives d ( + + d ( ( + + and then use the substitution u +, which gives d du ( ( + + (u +. Now, we use the trigonometric substitution u tan(θ: du (u + sec (θ (tan (θ + dθ sec (θ sec 4 (θ dθ dθ sec (θ cos (θdθ + cos(θdθ θ + sin(θ + D. 4 Net, we need to get back to the variable, so first, we use the identity: sin(θ sin(θ cos(θ, this gives θ + 4 sin(θ + D θ + sin(θ cos(θ + D. Now, since u tan(θ, we can draw a triangle to find both sin(θ and cos(θ, and we find that sin(θ and cos(θ. So u + u u + θ + sin(θ cos(θ + D tan (u + u u + + D. Finally, substituting u + gives du (u + tan ( ( D.

3 Section 7.4: 5. 0 ( ( + 9 d So partial fractions it is. We guess 0 ( ( + 9 A + B + C + 9, and can solve for A by setting. This gives A. So now the epression is B B + C C. Since this has to hold for all, this forces B (why? and hence C as well (again why?. So 0 ( ( + 9 d We now epand: d d d. + 9 d + 9 d. Each one of the above integrals can be done without too much trouble. So we have 0 ( ( + 9 d ln( ln( tan + R e e + 3e + d Here we set u e, and du e d, so we have e e + 3e + d e e + 3e + e d u u + 3u + du. The right-most integral above is in the form for partial fractions. So we set: u u + 3u + A u + + B u +, and find that B and A by setting u and u respectively. So now u u + 3u + du u + du ln(u + ln(u + + D. u + 3

4 After resubstitution, we get e e + 3e + d ln(e + ln(e + + D. 4

5 Part : The fun problems.. Consider the following integral: t 5 t + dt. a Solve the integral in two ways, first by using the substitution u tan(u and then solving the resulting trigonometric integral. Using the substitution suggested, we have dt sec (udu and t 5 t + dt ( 5 tan 5 (u sec tan (u + (udu ( tan 5 5 (u sec (u du sec(u ( 5 tan 5 (u sec(udu ( 5 tan 4 (u sec(u tan(udu ( 5 (sec (u sec(u tan(udu Now we are ready to set v sec(u and dv sec(u tan(udu. So we have ( (sec 5 (u sec(u tan(udu ( 5 (v dv ( 5 v 4 v + dv ( [ 5 5 v5 ] 3 v3 + v + D. Now we start back substituting. First up was that v sec(u. So ( [ 5 5 v5 ] 3 v3 + v + D ( [ 5 5 sec5 (u ] 3 sec3 (u + sec(u + D. And finally, we have to use the fact that t tan(u. Or more precisely, t 5

6 tan(u. Drawing a triangle gives that sec(u t +. So our final answer is [ t 5 t + dt ( 5 (t ( 5 3 ] (t + 3 ( t D 3 5 (t (t t + + D. b Solve the integral by the method of integration by parts. It might be helpful to rewrite the integral as follows: t t + dt, and let u t 4 and v t t +. So we follow the suggested first step. If u t 4 and v v t +. So we have t 4 t t + dt t4 t + 4t 3 t + dt t 4 t + 4 t t t + dt. t 4 t t +, then u 4t 3 and Notice that the integral on the second line has been put into the form for another substitution. Here, u t and v t t +, making u t and v 3 (t + 3. This gives t 4 t + 4 t 3 t + dt t 4 [ ] t t (t t(t + 3 dt t 4 t t (t t(t + 3 dt. Now the last integral is simple to do, so we now have t 4 t t + dt t4 t t (t (t D. c Your two answers, which should both be in terms of the original variable t. If you have done everything correctly, your answer to part a should look different than your answer to part b. Show that you answer to part a is indeed equal to your answer to part b. 6

7 We now have to compare the following two functions: f t 4 t t (t (t D and f 4 5 (t (t t + + D. The simplest way to do this is by direct calculation. I.e. we will factor a t + out of each function. After doing this, the first becomes t + (5 t 4 0 t ( t ( t + f and the second f Net, we notice that 5 ( t + 3 ( t + 0 t t 4 0 t ( t ( t + 3 ( t + 0 t + 0 3t 4 8t + 3. So finally, we have that f f t + ( 3t 4 8t Consider the following relationship between and z: z tan Show that under the above relation, one has {cos( z sin( +z z +z. Hint: Using double angle identities will help! So here, we use the hint: { cos( cos sin( sin cos. 7

8 Consider the first identity: cos( cos + tan ( Now the second: z + z. sin( sin cos tan z + z. sec ( + z sin ( cos sec ( tan cos + tan 3. Using problem, show the following: + cos( d tan + C Using problem, let z tan. Then cos( is given as above. However, we must compute d. Thus, one has + cos( d tan (z d + z + z +z dz dz + z. dz z + c tan + C. 8

1. Evaluate the integrals. a. (9 pts) x e x/2 dx. Solution: Using integration by parts, let u = x du = dx and dv = e x/2 dx v = 2e x/2.

1. Evaluate the integrals. a. (9 pts) x e x/2 dx. Solution: Using integration by parts, let u = x du = dx and dv = e x/2 dx v = 2e x/2. MATH 8 Test -SOLUTIONS Spring 4. Evaluate the integrals. a. (9 pts) e / Solution: Using integration y parts, let u = du = and dv = e / v = e /. Then e / = e / e / e / = e / + e / = e / 4e / + c MATH 8