Chapter 8

Calculate the integral: (a) by integrating by parts, (b) by applying a trigonometric substitution. $$\int x \arctan x d x$$

The integral of a rational function of \(\sinh x\) and \(\cosh x\) can be transformed into a rational function of \(u\) by means of the substitution \(u \quad \tanh \frac{1}{2} x .\) Show that this substitution gives $$\sinh x-\frac{2 u}{1-u^{2}} \cdot \quad \text { cosh } x=\frac{1+u^{2}}{1-u^{2}}, \quad d x=\frac{2}{1-u^{2}} d u$$

Calculate. $$\int_{1}^{2} x^{2}(\ln x)^{2} d x$$

Calculate using our table of integrals. $$\int \sqrt{4-x^{2}} d x$$

Derive the formula.$$\int \frac{d u}{u^{2}(a+b u)}=-\frac{1}{a u}+\frac{b}{a^{2}} \ln \left|\frac{a+b u}{u}\right|+C$$.

Calculate. (If you run out of ideas, use the examples as models.) $$\int_{0}^{2 / 4} \tan ^{3} x \sec ^{2} x d x$$.

Calculate the integral: (a) by integrating by parts, (b) by applying a trigonometric substitution. $$\int x \arcsin x d x$$

Integrate by setting \(u=\tanh \frac{1}{2} x.\) $$\int \operatorname{sech} x d x$$

$$\text { Derive }(8.2 .4): \int \ln x d x: x \ln x-x+C$$

Calculate using our table of integrals. $$\int \cos ^{3} 2 t d t$$

Derive the formula.$$\int \frac{d u}{u(a+b u)^{2}}=\frac{1}{a(a+b u)}+\frac{1}{a^{2}} \ln \left|\frac{a+b u}{u}\right|+C$$.

Calculate. (If you run out of ideas, use the examples as models.) $$\int_{\pi / 4}^{\pi / 2} \csc ^{3} x \cot x d x$$.

Find the area under the curve \(y=(\sqrt{x^{2}-9}) / x\) from \(x=3\) to \(x=5\)

Integrate by setting \(u=\tanh \frac{1}{2} x.\) $$\int \frac{1}{1+\cosh x} d x$$

Derive (8.2.6): $$\int \arctan x d x \quad x \arctan x-\frac{1}{2} \ln \left(1 ; x^{2}\right)+C$$

Derive the formula.$$\int \frac{d u}{a^{2}-u^{2}}=\frac{1}{2 a} \ln \left|\frac{a+u}{a-u}\right|+C$$.

Calculate. (If you run out of ideas, use the examples as models.) $$\int_{0}^{z / 6} \tan ^{2} 2 x d x$$.

The region under the curve \(y=1 /\left(1, x^{2}\right)\) from \(x=0\) to \(x=1\) is revolved about the \(x\) -axis. Find the volume of the resulting solid.

Integrate by setting \(u=\tanh \frac{1}{2} x.\) $$\int \frac{1}{\sinh x+\cosh x} d x$$

Derive the following three formulas. $$\int x^{k} \ln x d x=\frac{x^{k+1}}{k+1} \ln x-\frac{x^{k+1}}{(k+1)^{2}}+C, k \neq-1$$

Derive the formula.$$\int \frac{u d u}{a^{2}-u^{2}}=-\frac{1}{2} \ln \left|a^{2}-u^{2}\right|+C$$.

Calculate. (If you run out of ideas, use the examples as models.) $$\int_{0}^{\pi / 3} \tan x \sec ^{3 / 2} x d x$$.

Integrate by setting \(u=\tanh \frac{1}{2} x.\) $$\int \frac{1-e^{x}}{1+e^{x}} d x$$

Derive the following three formulas. $$\int e^{a x} \cos b x d x=\frac{e^{a x}(a \cos b x+b \sin b x)}{a^{2}+b^{2}}+C$$

Calculate using our table of integrals. $$\int \frac{x d x}{2+3 x}$$

Derive the formula.$$\int \frac{u^{2} d u}{a^{2}-u^{2}}=-u+\frac{a}{2} \ln \left|\frac{a+u}{a-u}\right|+C$$.

Find the ares benween the curve \(y=\sin ^{2} x\) and the \(x\) -axis from \(x=0\) to \(x=\pi\).

Show that in a disk of radius \(r\) a sector with central angle of radian measure \(\theta\) has area \(A=\frac{1}{2} r^{2} \theta .\) HINT: Assume first that \(0<\theta<\frac{1}{2} \pi\) and subdivide the region as indicated in the figure. Then verify that the formula holds for any sector.

Derive the following three formulas. $$\int e^{a x} \sin b x d x=\frac{e^{a x}(a \sin b x-b \cos b x)}{a^{2}+b^{2}}+C$$

Calculate using our table of integrals. $$\int \frac{\sqrt{x^{2}+9}}{x^{2}} d x$$

The region between the curve \(y=\cos x\) and the \(x\) -axis from \(x=-x / 2\) to \(x=\pi / 2\) is revolved about the \(x\) -axis. Find the volume of the resulting solid.

Find the area of the region bounded on the left and right by the two branches of the hyperbola \(\left(x^{2} / a^{2}\right)-\left(y^{2} / b^{2}\right)=1\) and above and below by the lines \(y=\pm b\)

Calculate.$$\int\left(u+\frac{d u}{b v)(c}, \cdot \cdot d \overline{u)}\right.$$,with the costficients as stipulated.a. \(b\). \(c . d\) all different trom \(0 . \quad a d=b c\).

Derive the following three formulas. What happens if you try integration by parts to calculate \(\int e^{a x} \cosh a x d x ?\) Calculate this integral by some other method.

Derive the following three formulas. Set \(f(x)=x \sin x .\) Find the area between the graph of \(f\) and the \(x\) -axis from \(x=0\) to \(x=\pi\)

Calculate.$$\int\left(u+\frac{d u}{b v)(c}, \cdot \cdot d \overline{u)}\right.$$,with the costficients as stipulated.a. \(b\). \(c . d\) all different from 0 , \(a d \neq b c\).

Find the area between the right branch of the hyperbola \(\left(x^{2} / 9\right)-\left(y^{2} / 16\right)=1\) and the line \(x=5\)

Show that for \(y=\frac{1}{x^{2}-1}\),$$\frac{d^{n} y}{d x^{n}}=\frac{(-1)^{n} n !}{2}\left[\frac{1}{(x-1)^{n+1}}-\frac{1}{(x+1)^{n+1}}\right]$$.

Tine region bounded by the \(y\) -axis and the curves \(y=\sin x\) and \(y=\cos x,=0: x \geq \pi / 4,\) is revolved about the \(x\) -axis. Find the volume of the resulting solid.

If the circle \((x-b)^{2}+y^{2}=a^{2}, b>a>0,\) is revolved about the \(y\) -axis, the resulting "doughnut-shaped" solid is called a torus. Use the shell method to find the formula for the volume of the torus.

Derive the following three formulas. Set \(g(x)=x \cos \frac{1}{2} x .\) Find the area between the graph of \(g\) and the \(x\) -axis from \(x=0\) to \(x=\pi\)

Calculate using our table of integrals. $$\int x^{3} \sin x d x$$

Find the volume of the solid generated by revolving the region between the curve \(y=1 / \sqrt{4-x^{2}}\) and the \(x\) -axis from \(x=0\) to \(x=3 / 2:\) (a) about the \(x\) -axis; (b) about the \(y\) -axis.

The region bounded by the \(y\) -axis, the line \(y=1\), and the curve \(y=\tan x \cdot x \in[0, \pi / 4],\) is revolved about the \(x\) -axis. Find the volume of the resulting solid.

Calculate the mass and the center of mass of a rod that extends from \(x=0\) to \(x=a>0\) and has mass density \(\lambda(x)=\left(x^{2}+a^{2}\right)^{-1 / 2}\)

Find the area between the graph of \(f\) and the \(x\) axis. $$f(x)=\arcsin x, \quad x \in\left[0, \frac{1}{2}\right]$$

Calculate using our table of integrals. $$\begin{aligned} &\text { Evaluate } \int_{0}^{ \pi} \sqrt{1+\cos x} d x\\\ &\mathrm{HINT}: \cos x=2 \cos ^{2} \frac{5}{2} x-1 \end{aligned}$$

The region between the curve \(y=\tan ^{2} x\) and the \(x\) -axis from \(x=0\) to \(x=\pi / 4\) is revolved about the \(x\) -axis. Find the volume of the resulting solid.

Find the area between the graph of \(f\) and the \(x\) axis. $$f(x)=x e^{-2 x}, \quad x \in[0,2]$$

Calculate \(\int \sec ^{2} x \tan x d x\) in two ways. (a) Sct \(\mu=\) lan \(x\) and verify that \(\int \sec ^{2} x \tan x d x=\frac{1}{2} \tan ^{2} x+c_{1}\) (b) Set \(u=\sec x\) and verify that \(\int \sec ^{2} x \tan x d x=\frac{1}{3} \sec ^{2} x+c_{2}\) (c) Reconcile the results in parts (a) and (b).