/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 5

Evaluate the following integrals: (i) \(\int \mathrm{e}^{\mathrm{x}}[\ln (\sec x+\tan \mathrm{x})+\sec \mathrm{x}] \mathrm{d} \mathrm{x}\) (ii) \(\int \mathrm{e}^{x}\left(\log x+\frac{1}{x^{2}}\right) d x\)

Short Answer

Expert verified
In conclusion, the step-by-step solutions for the given integrals are: (i) \(\int \mathrm{e}^{x}[\ln (\sec x+\tan x)+\sec x] dx = e^x \ln (\sec x+\tan x) + e^x \sec x - (x) - \int \frac{e^x \sec x}{\sec x + \tan x} dx\), where (x) represents an unsolvable integral. (ii) \(\int \mathrm{e}^{x}\left(\log x+\frac{1}{x^{2}}\right) dx = e^x \log x - Ei(x) + e^{-\frac{1}{x}} + C\), where \(Ei(x)\) is the exponential integral function, and C is the constant of integration.

Step by step solution

01

Choose u and dv

Let \(u=\ln(\sec x+\tan x)\) and \(dv=e^xdx\).
02

Find du and v

Compute \(du = \frac{\sec x}{\sec x + \tan x} dx\) by differentiating u, and \(v = e^x\) by integrating dv.
03

Apply integration by parts formula

Using the formula \(\int u dv = uv - \int v du\), we get: $$ \int e^x \ln (\sec x+\tan x) dx = e^x \ln (\sec x+\tan x) - \int \frac{e^x \sec x}{\sec x + \tan x} dx $$ The integral to the right is a difficult one to solve, so for now, we'll label it \((x)\) and move on to (b). (b) \(\int \mathrm{e}^{x} \sec x dx\)
04

Choose u and dv

Let \(u=\sec x\) and \(dv=e^xdx\).
05

Find du and v

Compute \(du = \sec x \tan x dx\) by differentiating u, and \(v = e^x\) by integrating dv.
06

Apply integration by parts formula

Using the formula \(\int u dv = uv - \int v du\), we get: $$ \int e^x \sec x dx = e^x \sec x - \int e^x \sec x \tan x dx $$ The integral to the right this time is solvable by substitution. Set \(y=\sec x+\tan x\), then \(dy=(\sec x \tan x + \sec^2x)dx\) Thus, $$ \int e^x \sec x \tan x dx = \int e^x dy = e^y = e^{\sec x + \tan x} $$ Putting it all together in (i): $$ \int \mathrm{e}^{x}[\ln (\sec x+\tan x)+\sec x] dx = e^x \ln (\sec x+\tan x) + e^x \sec x - (x) - \int \frac{e^x \sec x}{\sec x + \tan x} dx $$ Since we could not easily solve for (x), the result will be presented as the current expression. (ii) (a) \(\int \mathrm{e}^{x} \log x dx\)
07

Choose u and dv

Let \(u=\log x\) and \(dv=e^xdx\).
08

Find du and v

Compute \(du = \frac{1}{x} dx\) by differentiating u, and \(v = e^x\) by integrating dv.
09

Apply integration by parts formula

Using the formula \(\int u dv = uv - \int v du\), we get: $$ \int e^x \log x dx = e^x \log x - \int e^x \frac{1}{x} dx $$ There is no known elementary function that is the result of the integral on the right, so we can rewrite it as: $$ \int e^x \log x dx = e^x \log x - Ei(x) $$ where \(Ei(x)\) is the exponential integral function. (b) \(\int \mathrm{e}^x \frac{1}{x^2} dx\)
10

Choose substitution

Let \(y=-\frac{1}{x}\) and \(dy=\frac{1}{x^2} dx\).
11

Substitute and integrate

Substituting y, we get: $$ \int \mathrm{e}^x \frac{1}{x^2} dx = \int e^x dy = e^y = e^{-\frac{1}{x}} $$ Putting it all together in (ii): $$ \int \mathrm{e}^{x}\left(\log x+\frac{1}{x^{2}}\right) dx = e^x \log x - Ei(x) + e^{-\frac{1}{x}} + C $$ where C is the constant of integration.

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