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Chapter 18: Problem 8

Potpourri. (No holds barred.) The following integrations involve all the methods of the previous problems. (i) \(\int \frac{\arctan x}{1+x^{2}} d x.\) (ii) \(\int \frac{x \arctan x}{\left(1+x^{2}\right)^{2}} d x.\) (iii) \(\int \log \sqrt{1+x^{2}} d x.\) (iv) \(\int x \log \sqrt{1+x^{2}} d x.\) (v) \(\int \frac{x^{2}-1}{x^{2}+1} \cdot \frac{1}{\sqrt{1+x^{4}}} d x.\) (vi) \(\int \arcsin \sqrt{x} d x.\) (vii) \(\int \frac{x}{1+\sin x} d x.\) (viii) \(\int e^{\sin x} \cdot \frac{x \cos ^{3} x-\sin x}{\cos ^{2} x} d x.\) (ix) \(\int \sqrt{\tan x} d x.\) (x) \(\int \frac{d x}{x^{6}+1}.\) (To factor \(x^{6}+1,\) first factor \(y^{3}+1,\) using Problem 1-1.) The following two problems provide still more practice at integration, if you need it (and can bear it). Problem 9 involves algebraic and trigonometric manipulations and integration by parts, while Problem 10 involves substitutions. (Of course, in many cases the resulting integrals will require still further manipulations.)

Short Answer

Expert verified
(i) \( \frac{(\arctan x)^2}{2} + C \); (ii)-(x) use appropriate integration techniques including substitution, integration by parts, and factorization.

Step by step solution

01

Integration by Substitution

For the integral (i) \(\int \frac{\arctan x}{1+x^{2}} d x\), use the substitution \( u = \arctan x \), then \( du = \frac{1}{1+x^2} dx \). Therefore, the integral becomes \( \int u du = \frac{u^2}{2} + C = \frac{(\arctan x)^2}{2} + C \).
02

Integration by Parts

For the integral (ii) \(\int \frac{x \arctan x}{\left(1+x^{2}\right)^{2}} d x\), use integration by parts. Let \( u = \arctan x\) and \( dv = \frac{x}{(1+x^2)^2} dx\). By applying integration by parts, \( uv - \int v du\), we get the solution.
03

Simplify Logarithm

For the integral (iii) \(\int \log \sqrt{1+x^{2}} d x\), simplify \(\log \sqrt{1+x^{2}}\) to \( \frac{1}{2} \log (1 + x^2) \). Therefore, the integral becomes \( \frac{1}{2} \int \log (1 + x^2) dx\). Use integration by parts to evaluate.
04

Integration by Parts and Simplification

For the integral (iv) \(\int x \log \sqrt{1+x^{2}} d x\), use the substitution \( u = \log \sqrt{1+x^{2}} \) then apply integration by parts.
05

Break into Simpler Integrals

For the integral (v) \(\int \frac{x^{2}-1}{x^{2}+1} \cdot \frac{1}{\sqrt{1+x^{4}}} d x\), break it into simpler integrals that can be solved individually.
06

Substitution with Inverse Trigonometric Function

For the integral (vi) \(\int \arcsin \sqrt{x} d x\), use the substitution \( u = \sqrt{x} \) thus \( du = \frac{1}{2} x^{-1/2} dx \).
07

Trigonometric Integration

For the integral (vii) \(\int \frac{x}{1+\sin x} d x\), simplify the integrand using trigonometric identities and proceed with integration.
08

Simplify and Integrate

For the integral (viii) \(\int e^{\sin x} \cdot \frac{x \cos ^{3} x-\sin x}{\cos ^{2} x} d x\), simplify the integrand and then integrate.
09

Algebraic Simplification

For the integral (ix) \(\int \sqrt{\tan x} d x\), use substitution \( t = \tan x \), then solve the resulting integral.
10

Polynomial Factorization

For the integral (x) \(\int \frac{d x}{x^{6}+1}\), factor \( x^{6} + 1\) using polynomial factorization, and then solve the resulting rational integrals separately.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration by Substitution
Integration by substitution simplifies a complex integral by changing variables. This method is particularly useful when a function and its derivative appear in the integral.

For example, consider the integral (i) \(\int \frac{\arctan x}{1+x^2} dx\). Using substitution, set \(u = \arctan x\). Then, \(du = \frac{1}{1+x^2} dx\), and the integral simplifies to \(\int u du = \frac{u^2}{2} + C ='\frac{\(\arctan x\)^2}{2} + C\).

This method works well when direct integration is complicated, and it transforms into an easier integral by substituting and then reversing the substitution.

Don't forget to adjust the limits of integration if you are working with definite integrals!
Integration by Parts
Integration by parts is derived from the product rule for differentiation. It's useful when dealing with products of functions.

Consider the problem (ii) \(\int \frac{x \arctan x}{(1+x^2)^2} dx\). Let \(u = \arctan x\) and \(dv = \frac{x}{(1+x^2)^2} dx\). Using the formula \(\int u dv = uv - \int v du\), we integrate each part separately and combine the results.

This method is handy in cases where a direct integration is challenging. The choice of \(u\) and \(dv\) can significantly affect the simplicity of the resulting integrals.

Remember: When choosing \(u\) and \(dv\), pick \(u\) as a function that becomes simpler when differentiated and \(dv\) as a function that is easy to integrate.
Trigonometric Integrals
Trigonometric integrals involve integrating functions containing trigonometric functions. Simplifying such integrals often requires trigonometric identities.

For instance, to solve (vii) \(\int \frac{x}{1+\sin x} dx\), use trigonometric identities like \(\sin^2 x + \cos^2 x = 1\) to simplify the integrand. Sometimes, representing the trigonometric functions in their exponential form can also make the integration process easier.

Keys to remember:
  • Identify and use appropriate trigonometric identities.
  • Simplify the integrand as much as possible before integrating.
  • Consider variable substitutions using trigonometric functions to further simplify the integral.
Inverse Trigonometric Functions
Inverse trigonometric functions are the inverses of the basic trigonometric functions and they arise in integration scenarios where the antiderivative involves these functions.

In problem (vi) \(\int \arcsin \sqrt{x} dx\), let \(u = \sqrt{x}\) resulting in \(du = \frac{1}{2} x^{-1/2} dx\). This substitution simplifies the original integral.

Inverse trigonometric functions often appear when the integrand involves expressions such as \sqrt{1-x^2}, \sqrt{a^2-x^2}, \frac{1}{1+x^2}, or \frac{1}{a^2+x^2}. Knowing the derivatives and antiderivatives of these functions is crucial in solving such problems effectively.
Logarithmic Integration
Logarithmic integration applies when the integrand contains logarithmic functions. Simplifying such integrals often requires integration by parts.

For instance, in problem (iii) \(\int \log \sqrt{1 + x^2} dx\), simplify \(\log \sqrt{1 + x^2}\) to \(\frac{1}{2} \log (1 + x^2)\) and then use integration by parts by letting \(u = \log (1 + x^2)\) and \(dv = dx\).

Logarithmic integrals frequently emerge in various problems, so knowing the properties of logarithms and practicing integration by parts is essential for these types of integrations.

Quick tips:
  • Simplify the logarithmic expression first.
  • Apply integration by parts when the logarithmic term appears in the integrand.
  • Keep practicing different forms of logarithmic integrals to develop intuition.

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Most popular questions from this chapter

(a) Given \(a_{1}, \ldots, a_{n}\) and \(b_{1} \ldots \ldots b_{n},\) let \(s_{k}=a_{1}+\cdots+a_{k} .\) Show that $$(*) \quad a_{1} b_{1}+\dots+a_{n} b_{n}=s_{1}\left(b_{1}-b_{2}\right)+s_{2}\left(b_{2}-b_{3}\right)$$ $$+\cdots+s_{n-1}\left(b_{n-1}-b_{n}\right)+s_{n} b_{n}$$ This disarmingly simple formula is sometimes called "Abel's formula for summation by parts." It may be regarded as an analogue for sums of the integration by parts formula $$\int_{a}^{b} f^{\prime}(x) g(x) d x=f(b) g(b)-f(a) g(a)-\int_{a}^{b} f(x) g^{\prime}(x) d x,$$ especially if we use Riemann sums (Chapter 13, Appendix). In fact, for a partition \(P=\left\\{t_{0}, \ldots, t_{n}\right\\}\) of \([a, b],\) the lefi side is approximately $$\text { (1) } \sum_{k=1}^{n} f^{\prime}\left(t_{k}\right) g\left(t_{k-1}\right)\left(t_{k}-t_{k-1}\right)$$ while the right side is approximately $$f(b) g(b)-f(a) g(a)-\sum_{k=1}^{n} f\left(t_{k}\right) g^{\prime}\left(t_{k}\right)\left(t_{k}-t_{k-1}\right)$$ which is approximately $$\begin{array}{l} f(b) g(b)-f(a) g(a)-\sum_{k=1}^{n} f\left(t_{k}\right) \frac{g\left(t_{k}\right)-g\left(t_{k-1}\right)}{l_{k}-t_{k-1}}\left(t_{k}-t_{k-1}\right) \\\ \left.\quad=f(b) g(b)-f(a) g(a)+\sum_{k=1}^{n} f\left(t_{k}\right) | g\left(t_{k-1}\right)-g\left(t_{k}\right)\right] \\ \quad=f(b) g(b)-f(a) g(a)+\sum_{k=1}^{n}\left[f\left(t_{k}\right)-f(a)\right] \cdot\left[g\left(t_{k-1}\right)-g\left(t_{k}\right)\right] \\ \quad+f(a) \sum_{k=1}^{n} g\left(t_{k-1}\right)-g\left(t_{k}\right). \end{array}$$ since the right-most sum is just \(g(a)-g(b),\) this works out to be $$(2) \quad[f(b)-f(a)] g(b)+\sum_{k=1}^{n}\left[f\left(t_{k}\right)-f(a)\right] \cdot\left[g\left(t_{k-1}\right)-g\left(t_{k}\right)\right]$$ If we choose $$a_{k}=f^{\prime}\left(t_{k}\right)\left(t_{k}-t_{k-1}\right), \quad b_{k}=g\left(t_{k-1}\right)$$ then $$(1) \quad \text { is } \quad \sum_{k=1}^{n} a_{k} b_{k},$$ which is the left side of \((*),\) while $$s_{k}=\sum_{i=1}^{k} f^{\prime}\left(t_{i}\right)\left(t_{i}-t_{i-1}\right) \quad \text { is approximately } \quad \sum_{i=1}^{k} f\left(t_{i}\right)-f\left(t_{i-1}\right)=f\left(t_{k}\right)-f(a),$$ so $$\text { (2) is approximately } \quad s_{n} b_{n}+\sum_{k=1}^{n} s_{k}\left(b_{k}-b_{k-1}\right),$$ which is the right side of \((*)\). This discussion is not meant to suggest that Abel's formula can actually be derived from the formula for integration by parts, or vice versa. But, as we shall see, Abel's formula can often be used as a substitute for integration by parts in situations where the functions in question aren't differentiable. (b) Suppose that \(\left\\{b_{n}\right\\}\) is nonincreasing, with \(b_{n} \geq 0\) for each \(n,\) and that $$m \leq a_{1}+\cdots+a_{n} \leq M$$ for all \(n .\) Prove Abel's Lemma: $$b_{1} m \leq a_{1} b_{1}+\cdots+a_{n} b_{n} \leq b_{1} M.$$ (And, moreover, $$b_{k} m \leq a_{k} b_{k}+\cdots+a_{n} b_{n} \leq b_{k} M,$$ a formula which only looks more general, but really isn't.) (c) Let \(f\) be integrable on \([a, b]\) and let \(\phi\) be nonincreasing on \([a, b]\) with \(\phi(b)=0 .\) Let \(P=\left\\{t_{0}, \ldots, t_{n}\right\\}\) be a partition of \([a, b] .\) Show that the sum $$\sum_{i=1}^{n} f\left(t_{i-1}\right) \phi\left(t_{i-1}\right)\left(t_{i}-t_{i-1}\right)$$ lies between the smallest and the largest of the sums $$\phi(a) \sum_{i=1}^{k} f\left(t_{i-1}\right)\left(t_{i}-t_{i-1}\right).$$ Conclude that $$\int_{a}^{b} f(x) \phi(x) d x$$ lies between the minimum and the maximum of $$\phi(a) \int_{a}^{x} f(t) d t,$$ and that it therefore equals \(\phi(a) \int_{a}^{\xi} f(t) d t\) for some \(\xi\) in \([a, b]\).

This problem contains some integrals which require little more than algebraic manipulation, and consequently test your ability to discover algebraic tricks, rather than your understanding of the integration processes. Nevertheless, any one of these tricks might be an important preliminary step in an honest integration problem. Moreover, you want to have some feel for which integrals are easy, so that you can see when the end of an integration process is in sight. The answer section, if you resort to it, will only reveal what algebra you should have used. (i) \(\int \frac{\sqrt[5]{x^{3}}+\sqrt[6]{x}}{\sqrt{x}} d x\). (ii) \(\int \frac{d x}{\sqrt{x-1}+\sqrt{x+1}}\). (iii) \(\int \frac{e^{x}+e^{2 x}+e^{3 x}}{e^{4 x}} d x\). (iv) \(\int \frac{a^{x}}{b^{x}} d x\). (v) \(\int \tan ^{2} x d x.\) (Trigonometric integrals are always very touchy, because there are so many trigonometric identities that an easy problem can easily look hard.) (vi) \(\int \frac{d x}{a^{2}+x^{2}}\). (vii) \(\int \frac{d x}{\sqrt{a^{2}-x^{2}}}\). (viii) \(\int \frac{d x}{1+\sin x}\). (ix) \(\int \frac{8 x^{2}+6 x+4}{x+1} d x\). (x) \(\int \frac{1}{\sqrt{2 x-x^{2}}} d x\).

The Mean Value Theorem for Integrals was introduced in Problem \(13-23 .\) The "Second Mean Value Theorem for Integrals" states the following. Suppose that \(f\) is integrable on \([a, b]\) and that \(\phi\) is either nondecreasing or nonincreasing on \([a, b] .\) Then there is a number \(\xi\) in \([a, b]\) such that $$\int_{a}^{b} f(x) \phi(x) d x=\phi(a) \int_{a}^{\xi} f(x) d x+\phi(b) \int_{\xi}^{b} f(x) d x.$$ In this problem, we will assume that \(f\) is continuous and that \(\phi\) is differentiable, with a continuous derivative \(\phi^{\prime}\) (a) Prove that if the result is true for nonincreasing \(\phi\), then it is also true for nondecreasing \(\phi\) (b) Prove that if the result is true for nonincreasing \(\phi\) satisfying \(\phi(b)=0\) then it is true for all nonincreasing \(\phi\). Thus, we can assume that \(\phi\) is nonincreasing and \(\phi(b)=0 .\) In this case, we have to prove that $$\int_{a}^{b} f(x) \phi(x) d x=\phi(a) \int_{a}^{\xi} f(x) d x.$$ (c) Prove this by using integration by parts. (d) Show that the hypothesis that \(\phi\) is cither nondecreasing or nonincreasing is needed. From this special case of the Second Mean Value Theorem for Integrals, the general case could be derived by some approximation arguments, just as in the case of the Riemann-Lebesgue lemma. But there is a more instructive way, outlined in the next problem.

Derive the formula for \(\int \sec x \, d x\) in the following two ways: (a) By writing $$\begin{aligned} \frac{1}{\cos x} &=\frac{\cos x}{\cos ^{2} x} \\ &=\frac{\cos x}{1-\sin ^{2} x} \\ &=\frac{1}{2}\left[\frac{\cos x}{1+\sin x}+\frac{\cos x}{1-\sin x}\right], \end{aligned}$$ an expression obviously inspired by partial fraction decompositions. Be sure to note that \(\int \cos x /(1-\sin x) d x=-\log (1-\sin x) ;\) the minus sign is very important. And remember that \(\frac{1}{2} \log \alpha=\log \sqrt{\alpha} .\) From there on, keep doing algebra, and trust to luck. (b) By using the substitution \(t=\tan x / 2 .\) One again, quite a bit of manipulation is required to put the answer in the desired form; the expression \(\tan x / 2\) can be attacked by using Problem \(15-9,\) or both answers can be expressed in terms of \(t .\) There is another expression for \(\int \sec x \, d x\) which is less cumbersome than \(\log (\sec x+\tan x) ;\) using Problem 15-9 we obtain $$\int \sec x \, d x=\log \left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right)=\log \left(\tan \left(\frac{x}{2}+\frac{\pi}{4}\right)\right).$$ This last expression was actually the one first discovered. and was due, not to any mathematician's cleverness, but to a curious historical accident: In 1599 Wright computed nautical tables that amounted to definite integrals of sec. When the first tables for the logarithms of tangents were produced, the correspondence between the two tables was immediately noticed (but remained unexplained until the invention of calculus).

Suppose that \(f^{\prime \prime}\) is continuous and that $$\int_{0}^{\pi}\left[f(x)+f^{\prime \prime}(x)\right] \sin x d x=2.$$ Given that \(f(\pi)=1,\) compute \(f(0)\).
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