Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
8.1: 8 Evaluate the following integral using integration by parts: Z xe3x dx. We choose u to be x. Then we have: dv = e3x dx 1 v = e3x 3 u=x du = dx Applying the integration by parts formula gives: Z Z 1 1 3x 1 1 xe3x dx = xe3x − e = xe3x − e3x + C. 3 3 3 9 8.1: 68 Use the fomula Z f −1 (x) dx = xf −1 (x) − Z f (y) dy, (1) where y = f −1 (x), to evaluate the integral: Z tan−1 (x) dx. Express your answer in terms of x. By the formula, we have: Z tan−1 (x) dx = x tan−1 (x) − Z tan y dy = x tan−1 (x) − ln(cos y) + C = x tan−1 (x) − ln(cos(tan−1 x)) + C 1 = x tan−1 (x) + ln 1 + x2 + C 2 There are two steps in there that might require some explaining. To integrate tan y, rewrite tan y as sin y/ cos y, and make a u-substitution. And to simplify cos(tan−1 x), let θ = tan−1 x, and draw a reference triangle with the side opposite of θ having length x and the side adjacent to θ having length 1. Taking the cos of θ should then give you something similar to my result, though you’ll have to use ln rules to manipulate it to the form I’ve given. 1 8.2: 24 Evaluate the integral: π Z √ 1 − cos 2x dx 0 To evaluate this integral we’ll need to use the trig identity: 1 − cos 2θ . 2 Rearranging this gives something that looks quite similar to our integrand: √ √ 2 sin θ = 1 − cos 2θ. sin2 θ = Plugging this in, we see that: Z π √ 1 − cos 2x dx = √ Z 2 0 0 π π √ √ sin x dx = − 2 cos = 2 2. 0 8.2: 34 Evalute the integral: Z sec x tan2 x dx. Sometimes a small change can make a big difference - look at problem 33, which is ever so slightly different but can be done with a simple u-substitution of u = sec x. This one is not so simple. One way to go is to integrate by parts. Let: u = tan x dv = sec x tan x dx 2 du = sec x dx v = sec x Then applying the integration by parts formula, we find that: Z Z sec x tan2 x dx = sec x tan x − sec3 x dx. Here we use the fact that sec2 x = 1 + tan2 x. Then: Z Z Z sec x tan2 x dx = sec x tan x − sec x dx − sec x tan2 x dx. Notice now that the final integral on the right is what we started with. We can move this back over to the left to get: Z Z 2 sec x tan2 x dx = sec x tan x − sec x dx All that remains is to integrate sec x. To do this, multiply by 1 as follows: Z Z Z sec2 x + sec x tan x sec x + tan x dx = dx. sec x dx = sec x sec x + tan x sec x + tan x Notice now that the numerator is the derivative of the denominator, and so we can apply the u-substitution u = sec x + tan x. Putting everything together gives: Z 1 1 sec x tan2 x dx = sec x tan x − ln (sec x + tan x) . 2 2 2 8.3: 4 Evaluate the integral: 2 Z 1 dx. 8 + 2x2 0 We first put this into a better form for working with: Z 2 Z 1 1 2 1 dx = dx. 2 x 2 8 + 2x 8 0 0 1+ 2 x 2 We now make the substitution our limits!), we get: 1 8 Z 2 0 1 1+ = tan θ. Then dx = 2 sec2 θ dθ. Plugging in (and not forgetting to change dx = x 2 2 1 8 Z tan−1 (1) tan−1 (0) 1 1 2 sec2 θ dθ = 4 1 + tan2 θ Z π/4 dθ = 0 π . 16 8.3: 38 Use an appropriate substitution and then a trigonometric substitution to evaluate the integral: Z e 1 p dy. 2 1 y 1 + (ln y) We begin by letting x = ln y, which gives dx = e Z 1 1 y dy. Then substituting gives: Z 1 y p 1 + (ln y)2 dy = 1 √ 0 1 dx. 1 + x2 We now let x = tan θ, and dx = sec2 θ dθ. Then substitution gives: 1 Z √ 0 1 dx = 1 + x2 Z π/4 √ 0 1 1 + tan2 θ sec2 θ dθ. Simplifying gives an integral we’ve seen before in these solutions: Z 0 π/4 √ 1 1 + tan2 θ 2 Z sec θ dθ = 0 π/4 π/4 √ sec θ dθ = ln(sec θ + tan θ) = ln(1 + 2). 0 3