<p> Math 116 Exam 2 Fall 2014</p><p>Name:______This is a closed book exam. You may use a calculator and the formulas handed out with the exam. You may find that your calculator can do some of the problems. If this is so, you still need to show how to do the problem by hand. In other words, show all work and explain any reasoning which is not clear from the computations. (This is particularly important if I am to be able to give part credit.) Turn in this exam along with your answers. However, don't write your answers on the exam itself; leave them on the pages with your work. Also turn in the formulas; put them on the formula pile.</p><p>1. (24 points) Use integration by parts to find </p><p>2. (24 points) Find . Hints: Make a substitution to get rid of the sin's and cos's. Also the identity cos2x = 1 – sin2x should be helpful.</p><p>3. (16 points) Consider the integral . Make a trig substitution that gets rid of the square root and transforms the integral into an integral involving trig functions. You don't have to evaluate the resulting integral that you obtain after making the trig substituion.</p><p>4. (16 points) Complete the square and make a substitution that transforms the integral into one of the following forms: c or c or c where a and c are positive numbers you are to find. You don't have to evaluate the resulting integral that you obtain after completing the square and making the substitution. 5. (20 points) Use the partial fraction method to express as the sum of simpler fractions. Solutions 1. Let u = ln x and dv = x5 dx. So du = dx and v = = . } 8 points Therefore = ln x - = - . } 8 points Therefore = - . } 8 points 2. = } 4 points Let u = sin x. Then du = cos x dx and = } 10 points Expand the top and write the bottom as u1/2. = = = } 10 points 3. Let x = 4 sin u. So dx = 4 cos u du. Thus = = 4. = 2 . Complete the square with x2 - x + ¾. Divide coefficient of x by 2 and square. (-1/2)2 = ¼. Add and subtract this. x2 - x + ¾ = x2 - x + ¼ - ¼ + ¾ = (x – ½)2 + ½. So 2 = 2 . } 10 points Let u = x – ½. Then du = dx and 2 = 2 } 6 points 5. One has = + where A, B and C are to be determined. } 6 points Multiply by (x2 + 2)(x+1) to get 1 = (Ax + B)(x+1) + C(x2 + 2). } 2 points Plug in x = - 1  1 = C((- 1)2 + 2) = 3C  C = 1/3. Plug in x = 0  1 = B + 2C = B + 2(1/3)  B = 1/3. Plug in x = 1  1 = (A + B)(2) + C(3) = 2A + 2(1/3) + 3(1/3)  2A = 1 – 2/3 – 1 = - 2/3. So A = - 1/3. So, = + } 12 pts</p>