Solution. Every one is a substitution, and in each case the substitution is the inner function whose derivative already stands in the numerator, up to a constant.

(a)

\(u = 5-2x\), \(du = -2\,dx\): \(-\dfrac 12\displaystyle \int \frac {du}{u} = -\frac 12\ln \left |5-2x\right | + c\).

(b)

\(u = 3x^2\), \(du = 6x\,dx\): \(\dfrac 16\displaystyle \int e^u du = \frac {e^{3x^2}}{6} + c\).

(c)

\(u = 4x^3-5\), \(du = 12x^2dx\): \(\dfrac {1}{12}\displaystyle \int u^{-1/2}du = \frac {\sqrt {4x^3-5}}{6} + c\).

(d)

The derivative of the denominator is \(8x+4 = 4(2x+1)\), so with \(u = 4x^2+4x+3\), \(\dfrac 14\displaystyle \int \frac {du}{u} = \frac 14\ln \big (4x^2+4x+3\big ) + c\). No absolute value is needed: the quadratic has discriminant \(16-48<0\) and is always positive.

(e)

\(u = \ln \left |x\right |\), \(du = \dfrac {dx}{x}\): \(\displaystyle \int u\,du = \frac {\big (\ln \left |x\right |\big )^2}{2} + c\).

(f)

\(u = 2x^2+1\), \(du = 4x\,dx\): \(\dfrac 14\displaystyle \int \sqrt u\,du = \frac {\big (2x^2+1\big )^{3/2}}{6} + c\).

(g)

\(u = 1+\sin x\), \(du = \cos x\,dx\): \(\displaystyle \int \frac {du}{u} = \ln \left |1+\sin x\right | + c\).

(h)

\(u = \sin ^{-1}x\), \(du = \dfrac {dx}{\sqrt {1-x^2}}\): \(\displaystyle \int u\,du = \frac {\big (\sin ^{-1}x\big )^2}{2} + c\).

Solution.

(a)

Two applications, differentiating the polynomial each time: \[\int x^2\sin x\,dx = -x^2\cos x + 2\int x\cos x\,dx = -x^2\cos x + 2x\sin x + 2\cos x + c .\]

(b)

Take \(u = \cos ^{-1}x\), \(dv = dx\): \[x\cos ^{-1}x + \int \frac {x}{\sqrt {1-x^2}}dx = x\cos ^{-1}x - \sqrt {1-x^2} + c .\]

(c)

Applying parts twice returns the original integral. Writing \(I = \int e^x\sin x\,dx\), \[I = e^x\sin x - \int e^x\cos x\,dx = e^x\sin x - \Big [e^x\cos x + \int e^x\sin x\,dx\Big ] = e^x\big (\sin x - \cos x\big ) - I ,\] so \(2I = e^x(\sin x - \cos x)\) and \(I = \dfrac {e^x\big (\sin x-\cos x\big )}{2} + c\).

(d)

\(\displaystyle \int t^2e^{-t}dt = -\big (t^2+2t+2\big )e^{-t} + c\), again by two applications.

(e)

With \(u = (\ln x)^2\), \(dv = x\,dx\): \[\frac {x^2}{2}\big (\ln x\big )^2 - \int x\ln x\,dx = \frac {x^2}{2}\big (\ln x\big )^2 - \frac {x^2}{2}\ln x + \frac {x^2}{4} + c = \frac {x^2}{4}\Big [2\big (\ln x\big )^2 - 2\ln x + 1\Big ] + c .\]

(f)

With \(u = \tan ^{-1}x\), \(dv = x^2dx\): \[\frac {x^3}{3}\tan ^{-1}x - \frac 13\int \frac {x^3}{1+x^2}dx = \frac {x^3}{3}\tan ^{-1}x - \frac {x^2}{6} + \frac 16\ln \big (1+x^2\big ) + c ,\] using \(\dfrac {x^3}{1+x^2} = x - \dfrac {x}{1+x^2}\).

(g)

With the logarithm as \(u\) and \(dv = du\): \[u\ln \big (1+u^2\big ) - \int \frac {2u^2}{1+u^2}du = u\ln \big (1+u^2\big ) - 2u + 2\tan ^{-1}u + c .\]

Solution.

(a)

Write \(\cos ^nx = \cos ^{n-1}x\cdot \cos x\) and integrate by parts with \(u = \cos ^{n-1}x\), \(dv = \cos x\,dx\): \[\int \cos ^nx\,dx = \cos ^{n-1}x\sin x + (n-1)\int \cos ^{n-2}x\sin ^2x\,dx .\] Replacing \(\sin ^2x = 1-\cos ^2x\), \[= \cos ^{n-1}x\sin x + (n-1)\int \cos ^{n-2}x\,dx - (n-1)\int \cos ^nx\,dx .\] Collecting the original integral on the left gives \(n\int \cos ^nx\,dx\) equal to the stated right-hand side, and dividing by \(n\) finishes it.

(b)

Immediate from parts with \(u = x^n\), \(dv = e^xdx\).

(c)

Parts with \(u = (\ln x)^n\), \(dv = x^mdx\) gives \[\frac {x^{m+1}}{m+1}\big (\ln x\big )^n - \frac {n}{m+1}\int x^{m+1}\big (\ln x\big )^{n-1}\frac {dx}{x},\] and \(\dfrac {x^{m+1}}{x} = x^m\), which is the formula. Note the last exponent is \(n-1\), not \(n-2\) as printed in some copies.

The case \(n=2\), \(m=3\)

From (a): \(\displaystyle \int \cos ^2x\,dx = \frac {\cos x\sin x}{2} + \frac {x}{2} + c\).

From (b): \(\displaystyle \int x^2e^xdx = x^2e^x - 2\int xe^xdx = \big (x^2-2x+2\big )e^x + c\).

From (c) with \(m=3\), \(n=2\): \[\int x^3\big (\ln x\big )^2dx = \frac {x^4}{4}\big (\ln x\big )^2 - \frac 12\int x^3\ln x\,dx = \frac {x^4}{4}\big (\ln x\big )^2 - \frac {x^4}{8}\ln x + \frac {x^4}{32} + c .\]

Solution.

The reduction

Parts with \(u = (\ln x)^n\) and \(dv = dx\) gives \[I_n = x\big (\ln x\big )^n - \int x\cdot n\big (\ln x\big )^{n-1}\frac {dx}{x} = x\big (\ln x\big )^n - n\int \big (\ln x\big )^{n-1}dx = x\big (\ln x\big )^n - nI_{n-1} .\]

The definite integral

Starting from \(I_0 = x\) and applying the reduction, \[I_1 = x\ln x - x,\qquad I_2 = x\big (\ln x\big )^2 - 2\big (x\ln x - x\big ),\] \[I_3 = x\big (\ln x\big )^3 - 3\Big [x\big (\ln x\big )^2 - 2x\ln x + 2x\Big ].\] Evaluating between \(1\) and \(2\), and writing \(L = \ln 2\), the value at \(x=1\) is \(-3(0-0+2) = -6\) and at \(x=2\) it is \(2L^3 - 3\big (2L^2 - 4L + 4\big )\). Hence \[\int _1^2\big (\ln x\big )^3dx = 2L^3 - 6L^2 + 12L - 6 \approx 0.1345 ,\] where \(L = \ln 2\).

Solution. With \(x = \sin ^2 t\) we have \(dx = 2\sin t\cos t\,dt\) and \(1-x = \cos ^2t\), so \((1-x)^{3/2} = \cos ^3t\). The limits \(x=0,1\) become \(t = 0, \dfrac {\pi }{2}\), and \[\int _0^1x(1-x)^{3/2}dx = \int _0^{\pi /2}\sin ^2t\cdot \cos ^3t\cdot 2\sin t\cos t\,dt = 2\int _0^{\pi /2}\sin ^3t\cos ^4t\,dt .\] The power of \(\sin \) is odd, so substitute \(w = \cos t\): \[2\int _0^{\pi /2}\big (1-\cos ^2t\big )\cos ^4t\sin t\,dt = 2\int _0^1\big (1-w^2\big )w^4dw = 2\Big [\frac {w^5}{5} - \frac {w^7}{7}\Big ]_0^1 = 2\Big (\frac 15-\frac 17\Big ) = \frac {4}{35}.\]

Solution.

(a)

Differentiate the assumed form and match. Repeated use of the reduction \(\int x^ne^{3x}dx = \dfrac {x^ne^{3x}}{3} - \dfrac {n}{3}\int x^{n-1}e^{3x}dx\) gives \[a_0 = \frac 13,\quad a_1 = -\frac 49,\quad a_2 = \frac 49, \quad a_3 = -\frac {8}{27},\quad a_4 = \frac {8}{81}.\]

(b)

Using the reduction of the previous question with \(m=2\), \[a_0 = \frac 13,\quad a_1 = -\frac 13,\quad a_2 = \frac 29, \quad a_3 = -\frac {2}{27}.\] Both may be checked by differentiating the stated right-hand side and confirming the original integrand reappears.

Solution.

(a)

\(\sin ^2\theta = \dfrac {1-\cos 2\theta }{2}\), so the integral is \(\dfrac {\theta }{2} - \dfrac {\sin 2\theta }{4} + c\).

(b)

Both powers are even, so use double angles. Since \(\sin x\cos x = \dfrac {\sin 2x}{2}\), \[\sin ^2x\cos ^4x = \big (\sin x\cos x\big )^2\cos ^2x = \frac {\sin ^22x}{4}\cdot \frac {1+\cos 2x}{2} = \frac {\big (1-\cos 4x\big )\big (1+\cos 2x\big )}{16}.\] Expanding and using \(\cos 4x\cos 2x = \dfrac {\cos 6x + \cos 2x}{2}\), \[= \frac {1}{16}\Big [1 + \frac {\cos 2x}{2} - \cos 4x - \frac {\cos 6x}{2}\Big ],\] so the integral is \[\frac {x}{16} + \frac {\sin 2x}{64} - \frac {\sin 4x}{64} - \frac {\sin 6x}{192} + c .\]

(c)

\(\dfrac {1}{\sin ^26x} = \csc ^26x\), whose integral is \(-\dfrac {\cot 6x}{6} + c\).

(d)

\[\sin ^23x\sin ^25x = \frac {1-\cos 6x}{2}\cdot \frac {1-\cos 10x}{2} = \frac {1 - \cos 6x - \cos 10x + \cos 6x\cos 10x}{4},\] and \(\cos 6x\cos 10x = \dfrac {\cos 16x + \cos 4x}{2}\), so the integral is \[\frac {x}{4} - \frac {\sin 6x}{24} - \frac {\sin 10x}{40} + \frac {\sin 16x}{128} + \frac {\sin 4x}{32} + c .\]

(e)

\(\tan ^5x\cos x = \dfrac {\sin ^5x}{\cos ^4x}\). The power of \(\sin \) is odd, so with \(w = \cos x\), \[-\int \frac {\big (1-w^2\big )^2}{w^4}dw = -\int \Big (w^{-4} - 2w^{-2} + 1\Big )dw = \frac {1}{3w^3} - \frac {2}{w} - w + c ,\] that is \(\dfrac {1}{3\cos ^3x} - \dfrac {2}{\cos x} - \cos x + c\).

Solution.

(a)

Put \(x = 2\sin \theta \), so \(\sqrt {4-x^2} = 2\cos \theta \) and \(dx = 2\cos \theta \,d\theta \): \[\int \frac {2\cos \theta \,d\theta }{8\cos ^3\theta } = \frac 14\int \sec ^2\theta \,d\theta = \frac {\tan \theta }{4} = \frac {x}{4\sqrt {4-x^2}} + c .\]

(b)

Put \(x = 5\sin \theta \): \[\int \frac {5\cos \theta }{5\sin \theta }5\cos \theta \,d\theta = 5\int \frac {1-\sin ^2\theta }{\sin \theta }d\theta = 5\int \big (\csc \theta - \sin \theta \big )d\theta ,\] giving \(\sqrt {25-x^2} - 5\ln \left |\dfrac {5+\sqrt {25-x^2}}{x}\right | + c\).

(c)

Put \(x = 2\tan \theta \): \[\int \sqrt {x^2+4}\,dx = \frac {x\sqrt {x^2+4}}{2} + 2\ln \left |x+\sqrt {x^2+4}\right | + c .\]

(d)

Put \(x = 2\sec \theta \): \[\int \sqrt {x^2-4}\,dx = \frac {x\sqrt {x^2-4}}{2} - 2\ln \left |x+\sqrt {x^2-4}\right | + c .\]

(e)

Complete the square first: \(x^2-4x+13 = (x-2)^2+9\), so \[\int \frac {dx}{\sqrt {(x-2)^2+9}} = \ln \left |x-2+\sqrt {x^2-4x+13}\right | + c .\]

(f)

Again complete the square: \(4x-x^2 = 4-(x-2)^2\). With \(u = x-2\) this is part (a), giving \[\frac {x-2}{4\sqrt {4x-x^2}} + c .\]

Solution.

(a)

\(\dfrac {1}{x^2-9} = \dfrac {1}{6}\Big (\dfrac {1}{x-3} - \dfrac {1}{x+3}\Big )\), so the integral is \(\dfrac 16\ln \left |\dfrac {x-3}{x+3}\right | + c\).

(b)

The denominator factors as \(x(x-1)(x+2)\), and \[\frac {x^2-3x-1}{x(x-1)(x+2)} = \frac {1}{2x} - \frac {1}{x-1} + \frac {3}{2(x+2)},\] so the integral is \[\frac 12\ln \left |x\right | - \ln \left |x-1\right | + \frac 32\ln \left |x+2\right | + c .\]

(c)

The denominator is \((x-4)(x+1)\), and \(\dfrac {x}{(x-4)(x+1)} = \dfrac {4}{5(x-4)} + \dfrac {1}{5(x+1)}\), giving \[\frac 45\ln \left |x-4\right | + \frac 15\ln \left |x+1\right | + c .\]

(d)

The degrees are equal, so divide first: \[\frac {x^2+3x-4}{x^2-2x-8} = 1 + \frac {5x+4}{(x-4)(x+2)} = 1 + \frac {4}{x-4} + \frac {1}{x+2},\] and the integral is \(x + 4\ln \left |x-4\right | + \ln \left |x+2\right | + c\).

Solution.

(a)

Put \(x = w^2\), \(dx = 2w\,dw\): \[\int \frac {w}{1+w^2}2w\,dw = 2\int \frac {w^2}{1+w^2}dw = 2\int \Big (1 - \frac {1}{1+w^2}\Big )dw = 2w - 2\tan ^{-1}w ,\] that is \(2\sqrt x - 2\tan ^{-1}\sqrt x + c\).

(b)

Put \(w = \sqrt {x+2}\), so \(x = w^2-2\) and \(dx = 2w\,dw\): \[\int \frac {2w}{3+w}dw = 2\int \Big (1 - \frac {3}{3+w}\Big )dw = 2w - 6\ln \left |3+w\right | ,\] that is \(2\sqrt {x+2} - 6\ln \big (3+\sqrt {x+2}\big ) + c\).

(c)

Substitute \(w = \dfrac 1x\), so \(x = \dfrac 1w\) and \(dx = -\dfrac {dw}{w^2}\). Then \[x\sqrt {x^2+x-1} = \frac 1w\sqrt {\frac {1}{w^2} + \frac 1w - 1} = \frac {1}{w^2}\sqrt {1 + w - w^2}\quad (w>0),\] so the integral becomes \[-\int \frac {dw}{\sqrt {1+w-w^2}} = -\int \frac {dw}{\sqrt {\frac 54 - \big (w-\frac 12\big )^2}} = -\sin ^{-1}\!\left (\frac {2w-1}{\sqrt 5}\right ) + c ,\] that is \(-\sin ^{-1}\!\left (\dfrac {2-x}{x\sqrt 5}\right ) + c\).

(d)

Put \(x = 4\sin ^2\theta \), so \(\sqrt {4x-x^2} = 4\sin \theta \cos \theta \) and \(dx = 8\sin \theta \cos \theta \,d\theta \): \[\int \frac {4\sin \theta \cos \theta }{64\sin ^6\theta }8\sin \theta \cos \theta \,d\theta = \frac 12\int \frac {\cos ^2\theta }{\sin ^4\theta }d\theta = \frac 12\int \cot ^2\theta \csc ^2\theta \,d\theta = -\frac {\cot ^3\theta }{6}.\] Since \(\sin ^2\theta = \dfrac {x}{4}\) gives \(\cot \theta = \sqrt {\dfrac {4-x}{x}}\), the answer is \[-\frac 16\left (\frac {4-x}{x}\right )^{3/2} + c .\]

Solution. Parts (a), (b) and (d) are rational in \(\sin x\) and \(\cos x\), so use the Weierstrass substitution \(t = \tan \dfrac x2\), under which \[\sin x = \frac {2t}{1+t^2},\qquad \cos x = \frac {1-t^2}{1+t^2}, \qquad dx = \frac {2\,dt}{1+t^2}.\]

(a)

The denominator becomes \[1 + \frac {2t}{1+t^2} - \frac {1-t^2}{1+t^2} = \frac {2t^2+2t}{1+t^2},\] so the integral is \[\int \frac {1+t^2}{2t(t+1)}\cdot \frac {2\,dt}{1+t^2} = \int \frac {dt}{t(t+1)} = \ln \left |\frac {t}{t+1}\right | + c = \ln \left |\frac {\tan \frac x2}{1+\tan \frac x2}\right | + c .\]

(b)

The denominator becomes \(\dfrac {3+3t^2-2+2t^2}{1+t^2} = \dfrac {1+5t^2}{1+t^2}\), so \[\int \frac {2\,dt}{1+5t^2} = \frac {2}{\sqrt 5}\tan ^{-1}\big (\sqrt 5\,t\big ) + c = \frac {2}{\sqrt 5}\tan ^{-1}\!\Big (\sqrt 5\tan \frac x2\Big ) + c .\]

(c)

The standard trick is faster: multiply above and below by \(\sec x + \tan x\), whose derivative is the resulting numerator: \[\int \frac {\sec x\big (\sec x+\tan x\big )}{\sec x+\tan x}dx = \ln \left |\sec x + \tan x\right | + c .\]

(d)

The denominator becomes \(\dfrac {5+5t^2+8t}{1+t^2}\), so \[\int \frac {2\,dt}{5t^2+8t+5} = \int \frac {2\,dt}{5\Big [\big (t+\frac 45\big )^2 + \frac {9}{25}\Big ]} = \frac {2}{3}\tan ^{-1}\!\left (\frac {5t+4}{3}\right ) + c ,\] that is \(\dfrac 23\tan ^{-1}\!\left (\dfrac {5\tan \frac x2 + 4}{3}\right ) + c\).

Solution.

(a)

\(x^3\) is odd, so its integral over the symmetric interval is \(0\); \(2x^2\) is even, so \[\int _{-1}^{1}2x^2dx = 2\int _0^12x^2dx = \frac 43 .\] Noticing the symmetry avoids the work.

(b)

\(\Big [-2e^{-x/2}\Big ]_{-2}^{2} = -2e^{-1} + 2e = 2\Big (e - \dfrac 1e\Big ) \approx 4.70\).

(c)

The integrand is even, so this equals \(\displaystyle \int _3^5\sqrt {x^2-4}\,dx\). Using the antiderivative found above, \[\Big [\frac {x\sqrt {x^2-4}}{2} - 2\ln \left |x+\sqrt {x^2-4}\right |\Big ]_3^5 = \frac {5\sqrt {21}-3\sqrt 5}{2} - 2\ln \frac {5+\sqrt {21}}{3+\sqrt 5} \approx 6.894 .\]

(d)

With \(t = \tan \dfrac {\theta }{2}\), the denominator becomes \(\dfrac {5+5t^2+1-t^2}{1+t^2} = \dfrac {6+4t^2}{1+t^2}\), and \(\theta \) running from \(0\) to \(\dfrac {2\pi }{3}\) sends \(t\) from \(0\) to \(\sqrt 3\). So \[\int _0^{\sqrt 3}\frac {2\,dt}{6+4t^2} = \frac 12\int _0^{\sqrt 3}\frac {dt}{\frac 32+t^2} = \frac {1}{\sqrt 6}\tan ^{-1}\!\frac {t\sqrt 2}{\sqrt 3}\Big |_0^{\sqrt 3} = \frac {\tan ^{-1}\sqrt 2}{\sqrt 6}\approx 0.393 .\]

Solution.

(a)

The curves meet where \(9-x^2 = x+3\), that is \(x^2+x-6 = 0\), so \(x = -3\) and \(x = 2\). On that interval the parabola lies above the line, so \[A = \int _{-3}^{2}\Big [\big (9-x^2\big ) - \big (x+3\big )\Big ]dx = \int _{-3}^{2}\big (6 - x - x^2\big )dx = \Big [6x - \frac {x^2}{2} - \frac {x^3}{3}\Big ]_{-3}^{2} = \frac {125}{6}.\]

(b)

They meet where \(x^2-4 = 8-2x^2\), that is \(3x^2 = 12\) and \(x = \pm 2\). The second curve is the upper one, so \[A = \int _{-2}^{2}\Big [\big (8-2x^2\big )-\big (x^2-4\big )\Big ]dx = \int _{-2}^{2}\big (12-3x^2\big )dx = 32 .\]

Solution.

(a)

Discs of radius \(y\), with \(y^2 = x^2-16\), from the vertex \(x=4\) to \(x=8\): \[V = \pi \int _4^8\big (x^2-16\big )dx = \pi \Big [\frac {x^3}{3}-16x\Big ]_4^8 = \frac {256\pi }{3}.\]

(b)

Discs perpendicular to the \(y-\)axis of radius \(x\), with \(x^2 = \dfrac y4\): \[V = \pi \int _0^{16}\frac {y}{4}dy = \pi \Big [\frac {y^2}{8}\Big ]_0^{16} = 32\pi .\]

(c)

Now the axis is \(y = 16\), so the radius of a disc at abscissa \(x\) is \(16 - 4x^2\), and \(x\) runs from \(0\) to \(2\): \[V = \pi \int _0^2\big (16-4x^2\big )^2dx = \pi \int _0^2\big (256 - 128x^2 + 16x^4\big )dx = \frac {4096\pi }{15}.\]

(d)

Here \(y^2 = \Big [x^4\big (1-x^2\big )\Big ]^{2/3}\) and the loop runs from \(x=0\) to \(x=1\), so \[V = \pi \int _0^1x^{8/3}\big (1-x^2\big )^{2/3}dx .\] Substituting \(u = x^2\) turns this into a Beta integral, \[V = \frac {\pi }{2}\int _0^1u^{5/6}\big (1-u\big )^{2/3}du = \frac {\pi }{2}B\Big (\frac {11}{6},\frac 53\Big )\approx 0.401 .\] Unlike the others this does not reduce to elementary functions; the Beta function is the honest answer.

Solution.

(a)

\(y = 2x^{2/3}\), so \(y' = \dfrac {4}{3}x^{-1/3}\) and \[1 + (y')^2 = \frac {9x^{2/3}+16}{9x^{2/3}} .\] With \(u = 9x^{2/3}+16\), \(du = 6x^{-1/3}dx\), the integral becomes \(\dfrac {1}{18}\displaystyle \int \sqrt u\,du = \dfrac {u^{3/2}}{27}\), so \[L = \frac {1}{27}\Big [\big (9x^{2/3}+16\big )^{3/2}\Big ]_1^8 = \frac {52^{3/2}-125}{27} = \frac {104\sqrt {13}-125}{27}\approx 9.258 .\]

(b)

\(y = \dfrac {x^3}{6} + \dfrac {1}{2x}\), so \(y' = \dfrac {x^2}{2} - \dfrac {1}{2x^2}\) and \[1 + (y')^2 = \Big (\frac {x^2}{2} + \frac {1}{2x^2}\Big )^2 ,\] a perfect square, which is what makes these curves usable. Hence \[L = \int _1^2\Big (\frac {x^2}{2}+\frac {1}{2x^2}\Big )dx = \Big [\frac {x^3}{6} - \frac {1}{2x}\Big ]_1^2 = \frac {17}{12}.\]

(c)

\(y' = \dfrac 1x\), so \[L = \int _1^{2\sqrt 2}\frac {\sqrt {x^2+1}}{x}dx = \Big [\sqrt {x^2+1} - \ln \frac {1+\sqrt {x^2+1}}{x}\Big ]_1^{2\sqrt 2} = 3 - \sqrt 2 + \ln \frac {1+\sqrt 2}{\sqrt 2}\approx 2.121 .\]

(d)

\(\dot x = e^t(\cos t - \sin t)\) and \(\dot y = e^t(\sin t + \cos t)\), so \[\dot x^2 + \dot y^2 = e^{2t}\Big [(\cos t-\sin t)^2 + (\sin t+\cos t)^2\Big ] = 2e^{2t},\] and \[L = \int _0^4\sqrt 2\,e^t\,dt = \sqrt 2\big (e^4-1\big )\approx 75.8 .\]

Solution. About the \(x-\)axis the radius is \(y\); about the \(y-\)axis it is \(x\).

(a)

With \(y' = x^2\), \[S = 2\pi \int _0^3\frac {x^3}{3}\sqrt {1+x^4}\,dx .\] Putting \(u = 1+x^4\), \(du = 4x^3dx\), this is \(\dfrac {\pi }{6}\displaystyle \int _1^{82}\sqrt u\,du\), giving \[S = \frac {\pi }{9}\Big (82\sqrt {82}-1\Big )\approx 259.2 .\]

(b)

\[S = 2\pi \int _0^3x\sqrt {1+x^4}\,dx .\] With \(w = x^2\) this becomes \(\pi \displaystyle \int _0^9\sqrt {1+w^2}\,dw\), so \[S = \frac {\pi }{2}\Big (9\sqrt {82} + \sinh ^{-1}9\Big )\approx 132.6 .\]

(c)

The square root is again perfect, as in the arc length above: \[S = 2\pi \int _1^2x\Big (\frac {x^2}{2}+\frac {1}{2x^2}\Big )dx = 2\pi \int _1^2\Big (\frac {x^3}{2}+\frac {1}{2x}\Big )dx = \pi \Big (\frac {15}{4} + \ln 2\Big )\approx 13.96 .\]

(d)

With \(y' = \dfrac 1x\), \[S = 2\pi \int _1^7x\sqrt {1+\frac {1}{x^2}}\,dx = 2\pi \int _1^7\sqrt {x^2+1}\,dx = \pi \Big (34\sqrt 2 + \sinh ^{-1}7 - \ln \big (1+\sqrt 2\big )\Big )\approx 156.6 .\]