Calculus II by Paul Dawkins

By Paul Dawkins

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Extra info for Calculus II

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In other words we will need to set the coefficients of like powers of x equal. This will give a system of equations that can be solved. aspx Calculus II x3 : x2 : x1 : x0 : ü ï -4 A + B - 8C + D = 1 ï ý 3 A + 16C - 8D = -29 ï -12 A + 3B + 16 D = 5 ïþ A+C = 0 Þ A = 1, B = -5, C = -1, D = 2 Note that we used x0 to represent the constants. Also note that these systems can often be quite large and have a fair amount of work involved in solving them. The best way to deal with these is to use some form of computer aided solving techniques.

4æ xö æ 1 3 ö 2 ô cot ç ÷ dx = 2 ç - cot u - ò cot u du ÷ è2ø è 3 ø õ At this stage we can either reuse the reduction formula with n = 2 or use the formula ò cot 2 u du = - cot u - u + c We’ll reuse the reduction formula with n = 2 so we can address something that happens on occasion. æ 1 3 ó 4æ xö æ 1 öö 0 ô cot ç ÷ dx = 2 ç - cot u - ç - cot u - ò cot u du ÷ ÷ õ è2ø è 1 øø è 3 2 = - cot 3 u + 2 cot u + 2 ò du 3 2 = - cot 3 u + 2 cot u + 2u + c 3 2 æ xö æ xö = - cot 3 ç ÷ + 2 cot ç ÷ + x + c 3 è2ø è2ø cot 0 u = 1!

In these cases we really will need to use the second way of thinking about these kinds of terms. x3 Þ A B C + + x x 2 x3 x4 Þ A B C D + + + x x 2 x3 x 4 Let’s take a look at one more example. Example 6 Evaluate the following integral. 2 ó x dx ô 2 õ x -1 Solution In this case the numerator and denominator have the same degree. As with the last example we’ll need to do long division to get this into the correct form. I’ll leave the details of that to you to check. 2 ó x dx = ó 1 + 1 dx = dx + ó 1 dx ô ô 2 ò ôõ x 2 - 1 õ x2 - 1 õ x -1 So, we’ll need to partial fraction the second integral.

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