Chapter 7

In Problems 1-54, perform the indicated integrations. \(\int_{0}^{1} \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} d x\)

Explain what happens to the solution of the logistic differential equation if the initial population size is larger than the maximum capacity.

In Problems 37-48, apply integration by parts twice to evaluate each integral (see Examples 5 and 6). $$ \int \cos (\ln x) d x $$

In Problems 1-54, perform the indicated integrations. \(\int \frac{1}{x^{2}+2 x+5} d x\)

Without solving the logistic equation or referring to its solution, explain how you know that if \(y_{0}

In Problems 1-54, perform the indicated integrations. \(\int \frac{1}{x^{2}-4 x+9} d x\)

Find \(c\) so that \(\int_{0}^{c} \frac{1}{3 \sqrt{2 \pi}} x^{3 / 2} e^{-x / 2} d x=0.90\)

In Problems 1-54, perform the indicated integrations. \(\int \frac{d x}{9 x^{2}+18 x+10}\)

Suppose that the earth will not support a population of more than 16 billion and that there were 2 billion people in 1925 and 4 billion people in 1975 . Then, if \(y\) is the population \(t\) years after 1925 , an appropriate model is the logistic differential equation $$ \frac{d y}{d t}=k y(16-y) $$ (a) Solve this differential equation. (b) Predict the population in \(2015 .\) (c) When will the population be 9 billion?

Find \(c\) so that \(\int_{-c}^{c} \frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2} d x=0.95\). Hint: Use symmetry.

In Problems 1-54, perform the indicated integrations. \(\int \frac{d x}{\sqrt{16+6 x-x^{2}}}\)

In Problems 49-54, use integration by parts to derive the given formula. \(\begin{aligned} \int \cos 5 x \sin 7 x d x &=\\\ &-\frac{7}{24} \cos 5 x \cos 7 x-\frac{5}{24} \sin 5 x \sin 7 x+C \end{aligned}\)

In Problems 1-54, perform the indicated integrations. \(\int \frac{x+1}{9 x^{2}+18 x+10} d x\)

The Law of Mass Action in chemistry results in the differential equation $$ \frac{d x}{d t}=k(a-x)(b-x), \quad k>0, \quad a>0, \quad b>0 $$ where \(x\) is the amount of a substance at time \(t\) resulting from the reaction of two others. Assume that \(x=0\) when \(t=0\). (a) Solve this differential equation in the case \(b>a\). (b) Show that \(x \rightarrow a\) as \(t \rightarrow \infty\) (if \(b>a\) ). (c) Suppose that \(a=2\) and \(b=4\), and that 1 gram of the substance is formed in 20 minutes. How much will be present in 1 hour? (d) Solve the differential equation if \(a=b\).

In Problems 49-54, use integration by parts to derive the given formula. \(\int e^{\alpha z} \sin \beta z d z=\frac{e^{\alpha z}(\alpha \sin \beta z-\beta \cos \beta z)}{\alpha^{2}+\beta^{2}}+C\)

In Problems 1-54, perform the indicated integrations. \(\int \frac{3-x}{\sqrt{16+6 x-x^{2}}} d x\)

The differential equation $$ \frac{d y}{d t}=k(y-m)(M-y), y(0)=y_{0} $$ with \(k>0\) and \(0 \leq m

In Problems 49-54, use integration by parts to derive the given formula. \(\int e^{\alpha z} \cos \beta z d z=\frac{e^{\alpha z}(\alpha \cos \beta z+\beta \sin \beta z)}{\alpha^{2}+\beta^{2}}+C\)

In Problems 1-54, perform the indicated integrations. \(\int \frac{d t}{t \sqrt{2 t^{2}-9}}\)

As a model for the production of trypsin from trypsinogen in digestion, biochemists have proposed the model $$ \frac{d y}{d t}=k(A-y)(B+y) $$ where \(k>0, A\) is the initial amount of trypsinogen, and \(B\) is the original amount of trypsin. Solve this differential equation.

In Problems 49-54, use integration by parts to derive the given formula. \(\int x^{\alpha} \ln x d x=\frac{x^{\alpha+1}}{\alpha+1} \ln x-\frac{x^{\alpha+1}}{(\alpha+1)^{2}}+C, \alpha \neq-1\)

In Problems 1-54, perform the indicated integrations. \(\int \frac{\tan x}{\sqrt{\sec ^{2} x-4}} d x\)

Evaluate $$ \int_{\pi / 6}^{\pi / 2} \frac{\cos x}{\sin x\left(\sin ^{2} x+1\right)^{2}} d x $$

In Problems 49-54, use integration by parts to derive the given formula. \(\begin{aligned} \int x^{\alpha}(\ln x)^{2} d x &=\frac{x^{\alpha+1}}{\alpha+1}(\ln x)^{2} \\ &-2 \frac{x^{\alpha+1}}{(\alpha+1)^{2}} \ln x+2 \frac{x^{\alpha+1}}{(\alpha+1)^{3}}+C, \alpha \neq-1 \end{aligned}\)

Find the following derivatives. (a) \(\frac{d}{d x} \operatorname{erf}(x)\) (b) \(\frac{d}{d x} \operatorname{Si}(x)\)

Find the length of the curve \(y=\ln (\cos x)\) between \(x=0\) and \(x=\pi / 4\).

In Problems 55-61, derive the given reduction formula using integration by parts. $$ \int x^{\alpha} e^{\beta x} d x=\frac{x^{\alpha} e^{\beta x}}{\beta}-\frac{\alpha}{\beta} \int x^{\alpha-1} e^{\beta x} d x $$

Find the derivatives of the Fresnel functions (a) \(\frac{d}{d x} S(x)\) (b) \(\frac{d}{d x} C(x)\)

In Problems 55-61, derive the given reduction formula using integration by parts. $$ \int x^{\alpha} \sin \beta x d x=-\frac{x^{\alpha} \cos \beta x}{\beta}+\frac{\alpha}{\beta} \int x^{\alpha-1} \cos \beta x d x $$

Evaluate \(\int_{0}^{2 \pi} \frac{x|\sin x|}{1+\cos ^{2} x} d x\). Hint: Make the substitution \(u=x-\pi\) in the definite integral and then use symmetry properties.

In Problems 55-61, derive the given reduction formula using integration by parts. $$ \int x^{\alpha} \cos \beta x d x=\frac{x^{\alpha} \sin \beta x}{\beta}-\frac{\alpha}{\beta} \int x^{\alpha-1} \sin \beta x d x $$

In Problems 55-61, derive the given reduction formula using integration by parts. $$ \int(\ln x)^{\alpha} d x=x(\ln x)^{\alpha}-\alpha \int(\ln x)^{\alpha-1} d x $$

In Problems 55-61, derive the given reduction formula using integration by parts. $$ \begin{aligned} \int \cos ^{\alpha} \beta x d x &=\\\ & \frac{\cos ^{\alpha-1} \beta x \sin \beta x}{\alpha \beta}+\frac{\alpha-1}{\alpha} \int \cos ^{\alpha-2} \beta x d x \end{aligned} $$

Find the area of the region bounded by the curve \(y=\ln x\), the \(x\)-axis, and the line \(x=e\).

Find the area of the region bounded by the curves \(y=3 e^{-x / 3}, y=0, x=0\), and \(x=9\). Make a sketch.

Find the area of the region bounded by the graphs of \(y=x \sin x\) and \(y=x \cos x\) from \(x=0\) to \(x=\pi / 4\).

Find the volume of the solid obtained by revolving the region under the graph of \(y=\sin (x / 2)\) from \(x=0\) to \(x=2 \pi\) about the \(y\)-axis.

Evaluate the integral \(\int \cot x \csc ^{2} x d x\) by parts in two different ways: (b) By differentiating \(\csc x\) (a) By differentiating \(\cot x\) (c) Show that the two results are equivalent up to a constant.

If \(p(x)\) is a polynomial of degree \(n\) and \(G_{1}, G_{2}, \ldots, G_{n+1}\), are successive antiderivatives of a function \(g\), then, by repeated integration by parts, \(\int p(x) g(x) d x=p(x) G_{1}(x)-p^{\prime}(x) G_{2}(x)+p^{\prime \prime}(x) G_{3}(x)-\cdots\) \(+(-1)^{n} p^{(n)}(x) G_{n+1}(x)+C\) Use this result to find each of the following: (a) \(\int\left(x^{3}-2 x\right) e^{x} d x\) (b) \(\int\left(x^{2}-3 x+1\right) \sin x d x\)

The quantity \(a_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x d x\) plays an important role in applied mathematics. Show that if \(f^{\prime}(x)\) is continuous on \([-\pi, \pi]\), then \(\lim _{n \rightarrow \infty} a_{n}=0\). Hint: Integration by parts.

Find the error in the following "proof" that \(0=1\). In \(\int(1 / t) d t\), set \(u=1 / t\) and \(d v=d t\). Then \(d u=-t^{-2} d t\) and \(u v=1\). Integration by parts gives \(\int(1 / t) d t=1-\int(-1 / t) d t\)

Suppose that you want to evaluate the integral $$ \int e^{5 x}(4 \cos 7 x+6 \sin 7 x) d x $$ and you know from experience that the result will be of the form \(e^{5 x}\left(C_{1} \cos 7 x+C_{2} \sin 7 x\right)+C_{3}\). Compute \(C_{1}\) and \(C_{2}\) by differentiating the result and setting it equal to the integrand. Many surprising theoretical results can be derived through the use of integration by parts. In all cases, one starts with an integral. We explore two of these results here.

Show that $$ \begin{aligned} \int_{a}^{b} f(x) d x &=[x f(x)]_{a}^{b}-\int_{a}^{b} x f^{\prime}(x) d x \\ &=[(x-a) f(x)]_{a}^{b}-\int_{a}^{b}(x-a) f^{\prime}(x) d x \end{aligned} $$

Show that $$ f(t)=f(a)+\sum_{i=1}^{n} \frac{f^{(i)}(a)}{i !}(t-a)^{i}+\int_{a}^{t} \frac{(t-x)^{n}}{n !} f^{(n+1)}(x) d x, $$ provided that \(f\) can be differentiated \(n+1\) times.

The Beta function, which is important in many branches of mathematics, is defined as $$ B(\alpha, \beta)=\int_{0}^{1} x^{\alpha-1}(1-x)^{\beta-1} d x $$ with the condition that \(\alpha \geq 1\) and \(\beta \geq 1\). (a) Show by a change of variables that $$ B(\alpha, \beta)=\int_{0}^{1} x^{\beta-1}(1-x)^{\alpha-1} d x=B(\beta, \alpha) $$ (b) Integrate by parts to show that \(B(\alpha, \beta)=\frac{\alpha-1}{\beta} B(\alpha-1, \beta+1)=\frac{\beta-1}{\alpha} B(\alpha+1, \beta-1)\) (c) Assume now that \(\alpha=n\) and \(\beta=m\), and that \(n\) and \(m\) are positive integers. By using the result in part (b) repeatedly, show that $$ B(n, m)=\frac{(n-1) !(m-1) !}{(n+m-1) !} $$ This result is valid even in the case where \(n\) and \(m\) are not integers, provided that we can give meaning to \((n-1) !\), \((m-1)\) !, and \((n+m-1) !\).

Suppose that \(f(t)\) has the property that \(f^{\prime}(a)=f^{\prime}(b)=0\) and that \(f(t)\) has two continuous derivatives. Use integration by parts to prove that \(\int_{a}^{b} f^{\prime \prime}(t) f(t) d t \leq 0\). Hint: Use integration by parts by differentiating \(f(t)\) and integrating \(f^{\prime \prime}(t)\). This result has many applications in the field of applied mathematics.

Derive the formula $$ \int_{0}^{x}\left(\int_{0}^{t} f(z) d z\right) d t=\int_{0}^{x} f(t)(x-t) d t $$ using integration by parts.