18.01 Solutions, Problem Set 8

• §9.5 #14.

If a is a positive constant, show that

∫ \frac{du}{\sqrt{a^{2} − u^{2}}} = sin^{−1} \frac{u}{a} + C

and

∫ \frac{du}{a^{2} + u^{2}} = \frac{1}{a} tan^{−1} \frac{u}{a} + C.

Solution: Since \frac{d}{dx} sin^{−1} x = 1/\sqrt{1 − x^{2}},

\frac{d}{du} sin^{−1}(u/a) = \frac{1}{\sqrt{1 − (u/a)^{2}}} · \frac{d}{du}(u/a)
= \frac{1}{a\sqrt{1 − u^{2}/a^{2}}}
= \frac{1}{\sqrt{a^{2}(1 − u^{2}/a^{2})}}
= \frac{1}{\sqrt{a^{2}(1 − u^{2}/a^{2})}}
= \frac{1}{\sqrt{a^{2} − u^{2}}}
so ∫ \frac{du}{\sqrt{a^{2} − u^{2}}} = sin^{−1} \frac{u}{a} + C.

Since \frac{d}{dx} tan^{−1} x = 1/(1 + x^{2}),

\frac{d}{du} (\frac{1}{a} tan^{−1}(u/a)) = \frac{1}{a(1 + (u/a)^{2})} · \frac{d}{du}(u/a)
= \frac{1}{a^{2}(1 + u^{2}/a^{2})}
= \frac{1}{a^{2} + u^{2}}
so ∫ \frac{du}{a^{2} + u^{2}} = \frac{1}{a} tan^{−1} \frac{u}{a} + C.

• §9.5 #16.

Evaluate ∫_{−1}^{1} \frac{dx}{1 + x^{2}} .

Solution:

∫_{−1}^{1} \frac{dx}{1 + x^{2}} = tan^{−1} x]_{−1}^{1}
= tan^{−1}(1) − tan^{−1}(−1)
= π/4 − (−π/4)
= π/2.

• §9.5 #17. Evaluate ∫ \frac{dx}{\sqrt{1 − 4x^{2}}} .

Solution: The substitution u = 2x, du = 2 dx gives

∫ \frac{dx}{\sqrt{1 − 4x^{2}}} = ∫ \frac{2 dx}{2\sqrt{1 − (2x)^{2}}}
= ∫ \frac{du}{2\sqrt{1 − u^{2}}}
= \frac{1}{2} sin^{−1} u + C
= \frac{1}{2} sin^{−1}(2x) + C.

• §9.7 #13.

Find \frac{dy}{dx} if y = sinh(x^{3}).

Solution: Since \frac{d}{du} sinh u = cosh u, the chain rule gives

\frac{d}{dx} sinh(x^{3}) = cosh(x^{3})\frac{d}{dx}(x^{3}) = 3x^{2} cosh(x^{3}).

• §9.7 #15.

Find \frac{dy}{dx} if y = ln(tanh(3x)).

Solution: Since \frac{d}{du} tanh u = sech^{2}u, the chain rule gives

\frac{d}{dx} ln tanh(3x) = \frac{1}{tanh(3x)} \frac{d}{dx} tanh(3x)
= \frac{1}{tanh(3x)} · 3sech^{2}(3x)
= \frac{3 cosh(3x)}{sinh(3x) cosh^{2}(3x)}
= 3csch(3x)sech(3x)

Also, note that sinh(2u) = 2 sinh(u) cosh(u) carries over without a sign change from the regular trigonometric. Thus

3csch(3x)sech(3x) = 6csch(6x)

as listed in the solutions in the back of the textbook.

• §9.7 #20.

Find ∫ \frac{sinh(3x)}{1 + cosh(3x)} dx.

Solution: Substitute u = 1 + cosh(3x), du = 3 sinh(3x), so

\frac{1}{3} ∫ \frac{3 sinh(3x)}{1 + cosh(3x)} dx = \frac{1}{3} ∫ \frac{du}{u}
= \frac{1}{3} ln |u| + C
= \frac{1}{3} ln(1 + cosh(3x)) + C.

• §10.2 Find the following integrals:

#19. ∫ \frac{e^{tan^{−1} x}}{1 + x^{2}} dx.

Solution: Substitute u = tan^{−1} x, du = \frac{dx}{1 + x^{2}} :

∫ \frac{e^{tan^{−1} x}}{1 + x^{2}} dx = ∫ e^{u} du = e^{u} + C = e^{tan^{−1} x} + C.

#27. ∫ \frac{e^{x}}{\sqrt{1 − e^{2x}}} dx.

Solution: Substitute u = e^{x}, du = e^{x} dx:

∫ \frac{e^{x}}{\sqrt{1 − (e^{x})^{2}}} dx = ∫ \frac{1}{\sqrt{1 − u^{2}}} du = sin^{−1} u + C = sin^{−1}(e^{x}) + C.

#38. ∫ \frac{sin^{−1} x}{\sqrt{1 − x^{2}}} dx.

Solution: Substitute u = sin^{−1} x, du = \frac{1}{\sqrt{1 − x^{2}}} dx:

∫ \frac{sin^{−1} x}{\sqrt{1 − x^{2}}} dx = ∫ u du = \frac{1}{2}u^{2} + C = \frac{1}{2}(sin^{−1} x)^{2} + C.

#45. ∫_{1}^{2} \frac{(2x + 1) dx}{\sqrt{x^{2} + x + 2}}.

Solution: Substitute u = x^{2} +x+2, du = (2x+1) dx (note the new integral is from u = 1^{2} + 1 + 2 = 4 to u = 2^{2} + 2 + 2 = 8):

∫_{1}^{2} \frac{(2x + 1) dx}{\sqrt{x^{2} + x + 2}} = ∫_{4}^{8} u^{−1/2} du = 2u^{1/2}]_{4}^{8} = 4(\sqrt{2} − 1).

#48. ∫_{0}^{π/3} sec^{3} x tan x dx.

Solution: Substitute u = sec x, du = sec x tan x dx (the new integral is from u = sec 0 = 1 to u = sec(π/3) = 2)

∫_{0}^{π/3} sec^{2} x sec x tan x dx = ∫_{1}^{2} u^{2} du = \frac{1}{3}u^{3}]_{1}^{2} = \frac{7}{3}.

• §10.3 Find the following integrals:

#2. ∫ sin^{4} x dx.

Solution: The formula sin^{2} x = \frac{1}{2}(1 − cos(2x)) gives

sin^{4} x = \frac{1}{4}(1 − cos(2x))^{2} = \frac{1}{4}(1 − 2 cos(2x) + cos^{2}(2x)).

Since cos^{2}(2x) = \frac{1}{2}(1 + cos(4x)), this becomes

\frac{1}{4}(1 − 2 cos(2x) + \frac{1}{2}(1 + cos(4x))) = \frac{1}{8}(3 − 4 cos(2x) + cos(4x)).

Since ∫ cos(ax) dx = \frac{1}{a} sin(ax) + C, we get

∫ sin^{4} x dx = \frac{1}{8} ∫ (3 − 4 cos(2x) + cos(4x)) dx
= \frac{3}{8}x − \frac{1}{4} sin(2x) + \frac{1}{32} sin(4x) + C.

#6. ∫ sin^{2} x cos^{5} x dx.

Solution: The substitution u = sin x, du = cos x dx, together with the identity cos^{2} x = 1 − sin^{2} x gives

∫ sin^{2} x cos^{5} x dx = ∫ sin^{2} x(cos^{2} x)^{2} cos x dx
= ∫ u^{2}(1 − u^{2})^{2} du

= ∫ u^{6} − 2u^{4} + u^{2} du
= \frac{1}{7}u^{7} − \frac{2}{5}u^{5} + \frac{1}{3}u^{3} + C
= \frac{1}{7} sin^{7} x − \frac{2}{5} sin^{5} x + \frac{1}{3} sin^{3} x + C.

#9. ∫ \sqrt{sin x} cos^{3} x dx.

Solution: Substituting u = sin x and du = cos x dx gives

∫ \sqrt{sin x} cos^{3} x dx = ∫ (sin x)^{1/2}(1 − sin^{2}x) cos x dx
= ∫ u^{1/2}(1 − u^{2}) du
= ∫ (u^{1/2} − u^{5/2}) du
= \frac{2}{3}u^{3/2} − \frac{2}{7}u^{7/2} + C
= (\frac{2}{3} sin x − \frac{2}{7} sin^{3} x) \sqrt{sin x} + C.

#13. ∫_{0}^{π/4} sec^{4} x dx.

Solution: Substituting u = tan x, du = sec^{2} x dx and using the identity sec^{2} x = 1 + tan^{2} x (note the new limits of integration) gives:

∫_{0}^{π/4} sec^{4} x dx = ∫_{0}^{π/4}(1 + tan^{2} x) sec^{2} x dx
= ∫_{0}^{1}(1 + u^{2}) du = (u + \frac{1}{3}u^{3})]_{0}^{1} = \frac{4}{3}.

• §10.3 #26.

If n is any positive integer ≥ 2, show that

∫ tan^{n} x dx = \frac{tan^{n−1} x}{n − 1} − ∫ tan^{n−2} x dx.

Solution: The identity tan^{2} x = sec^{2} x−1 gives tan^{n} x = tan^{n−2} x sec^{2} x− tan^{n−2}, so

∫ tan^{n} x dx = ∫ tan^{n−2} x sec^{2} x dx − ∫ tan^{n−2} x dx.

Evaluating the first integral on the right-hand-side using the substitution u = tan x, du = sec^{2} x dx, we get

∫ tan^{n−2} x sec^{2} x dx = ∫ u^{n−2} du = \frac{u^{n−1}}{n − 1} + C = \frac{tan^{n−1} x}{n − 1} + C.

(We can “absorb” the constant in the second indefinite integral above.)

• §10.4 Find the following integrals:

#2. ∫ \frac{x^{2} dx}{\sqrt{4 − x^{2}}} .

Solution: We substitute x = 2 sin θ, so θ = sin^{−1} \frac{x}{2}, dx = 2 cos θ dθ

and \sqrt{4 − x^{2}} = 2 cos θ. This gives

∫ \frac{x^{2} dx}{\sqrt{4 − x^{2}}} = ∫ \frac{4 sin^{2} θ · 2 cos θ dθ}{2 cos θ}
= ∫ 4 sin^{2} θ, dθ
= ∫ (2 − 2 cos(2θ)) dθ
= 2θ − sin(2θ) + C.

To express sin(2θ) in terms of x, we could use the identity sin(2θ) = 2 sin θ cos θ, which gives

2θ − sin(2θ) + C = 2 sin^{−1} \frac{x}{2} − \frac{x}{2}\sqrt{4 − x^{2}} + C.

#6. ∫ \frac{dx}{x\sqrt{a^{2} − x^{2}}} .

Solution: We substitute x = a sin θ, so θ = sin^{−1} \frac{x}{a}, dx = a cos θ dθ

and \sqrt{a^{2} − x^{2}} = a cos θ. This gives

∫ \frac{dx}{x\sqrt{a^{2} − x^{2}}} = ∫ \frac{a cos θ dθ}{a^{2} sin θ cos θ}
= ∫ \frac{1}{a} csc θ dθ
= −\frac{1}{a} ln | csc θ + cot θ| + C
= −\frac{1}{a} ln |\frac{1 + cos θ}{sin θ} | + C

= −\frac{1}{a} ln |\frac{1 + \frac{1}{a} \sqrt{a^{2} − x^{2}}}{\frac{x}{a}} | + C

= −\frac{1}{a} ln |\frac{a + \sqrt{a^{2} − x^{2}}}{x} | + C.

#14. ∫ \frac{dx}{(a^{2} − x^{2})^{3/2}} .

Solution: Again we use x = a sin θ, so θ = sin^{−1} \frac{x}{a}, dx = a cos θ dθ

and \sqrt{a^{2} − x^{2}} = a cos θ. This gives

∫ \frac{dx}{(a^{2} − x^{2})^{3/2}} = ∫ \frac{a cos θ dθ}{a^{3} cos^{3} θ}
= ∫ \frac{1}{a^{2}} sec^{2} θ dθ
= \frac{1}{a^{2}} tan θ + C
= \frac{sin θ}{a^{2} cos θ} + C
= \frac{x}{a^{2}\sqrt{a^{2} − x^{2}}} + C

#18. ∫ \frac{dx}{(x^{2} − a^{2})^{3/2}} .

Solution: Now we use x = a sec θ, θ = sec^{−1} \frac{x}{a}, dx = a sec θ tan θ dθ

and \sqrt{x^{2} − a^{2}} = a tan θ. This gives

∫ \frac{dx}{(x^{2} − a^{2})^{3/2}} = ∫ \frac{a sec θ tan θ dθ}{a^{3} tan^{3} θ} = \frac{1}{a^{2}} ∫ \frac{cos θ dθ}{sin^{2} θ} .

Expressing the integrand this way suggests the substitution u = sin θ, du = cos θ dθ, which gives

\frac{1}{a^{2}} ∫ \frac{cos θ dθ}{sin^{2} θ} = \frac{1}{a^{2}} ∫ u^{−2} du = \frac{−1}{ua^{2}} + C = \frac{−1}{a^{2} sin θ} + C.

We just have to express sin θ in terms of x. Reading this off from the triangle with side-lengths a, x = a sec θ and \sqrt{x^{2} − a^{2}} = a tan θ gives sin θ = (a tan θ)/(a sec θ) = \sqrt{x^{2} − a^{2}}/x, so

∫ \frac{dx}{(x^{2} − a^{2})^{3/2}} = \frac{−x}{a^{2}\sqrt{x^{2} − a^{2}}} + C.

• §10.4 #31. If the circle (x − b)^{2} + y^{2} = a^{2} (0 < a < b) is revolved about the y-axis, the resulting solid of revolution is called a torus. Use the shell method to find the volume of this torus.

Solution: We integrate dV = 2πxh dx from x = b − a to x = b + a, where h is the height of the cylindrical shell, so h = 2\sqrt{a^{2} − (x − b)^{2}}. So

V = 4π ∫_{b−a}^{b+a} x\sqrt{a^{2} − (x − b)^{2}} dx.

This suggests substituting so that x − b = a sin θ, i.e., x = b + a sin θ, so

θ = sin^{−1} \frac{x − b}{a}, dx = a cos θ dθ and \sqrt{a^{2} − (x − b)^{2}} = a cos θ. This gives

V = 4π ∫_{−π/2}^{π/2}(b + a sin θ)a^{2} cos^{2} θ dθ

= 4πa^{2}b ∫_{−π/2}^{π/2} cos^{2} θ dθ + 4πa^{3} ∫_{−π/2}^{π/2} sin θ cos^{2} θ dθ.

We evaluate the first integral using the identity cos^{2} θ = \frac{1}{2}(1 + cos(2θ)), giving

πa^{2}b ∫_{−π/2}^{π/2}(1 + cos(2θ)) 2dθ = πa^{2}b ∫_{−π}^{π}(1 + cos u) du
= πa^{2}b(u + sin u)]_{−π}^{π} = 2π^{2}a^{2}b.

For the second integral, we substitute v = cos θ, dv = − sin θ dθ, so

4πa^{3} ∫_{−π/2}^{π/2} sin θ cos^{2} θ dθ = 4πa^{3} ∫_{0}^{0} −v^{2} dv = 0.

Therefore the volume V is 2π^{2}a^{2}b.

• §10.4 #33. Find the length of the curve y = ln x between x = 1 and x = \sqrt{8}.

Solution: We integrate

ds = \sqrt{1 + (dy/dx)^{2}} dx = \sqrt{1 + \frac{1}{x^{2}}} dx = \sqrt{\frac{x^{2} + 1}{x^{2}}} dx = \frac{\sqrt{x^{2} + 1}}{x} dx

from x = 1 to x = \sqrt{8}. We first find the indefinite integral using the substitution x = tan θ, so θ = tan^{−1} x, dx = sec^{2} θ dθ and \sqrt{x^{2} + 1} = sec θ. This gives

∫ \frac{\sqrt{x^{2} + 1}}{x} dx = ∫ \frac{sec θ · sec^{2} θ dθ}{tan θ}
= ∫ \frac{sec θ(tan^{2} θ + 1) dθ}{tan θ}
= ∫ (sec θ tan θ + csc θ) dθ
= sec θ − ln | csc θ + cot θ| + C.

To express this in terms of x, use csc θ = \frac{sec θ}{tan θ} and cot θ = \frac{1}{tan θ}. This gives:

∫ \frac{\sqrt{x^{2} + 1}}{x} dx == \sqrt{x^{2} + 1} − ln |\frac{\sqrt{x^{2} + 1} + 1}{x}| + C.

Subtracting the value at x = 1 from the value at x = \sqrt{8} gives

s = 3 − ln \sqrt{2} − \sqrt{2} + ln(\sqrt{2} + 1) = 3 − \sqrt{2} + ln(1 + \sqrt{2}/2).

• §10.5 Calculate the following integrals:

#2. ∫ \frac{dx}{\sqrt{5 + 4x − x^{2}}} .

Solution: We complete the square by writing 5 + 4x − x^{2} = 5 + 4 − 4 + 4x − x^{2} = 9 − (4 − 4x + x^{2}) = 9 − (x − 2)^{2}. The substitution u = x − 2, du = dx gives

∫ dx\sqrt{9 − (x − 2)^{2}} = ∫ du\sqrt{9 − u^{2}} = sin^{−1} \frac{u}{3}+C = sin^{−1} \frac{x − 2}{3}+C.

#7. ∫ \frac{x^{2} dx}{\sqrt{6x − x^{2}}} .

Solution: Complete the square by writing

6x − x^{2} = 9 − 9 + 6x − x^{2} = 3^{2} − (x − 3)^{2}.

Now substitute so that x − 3 = 3 sin θ, i.e., x = 3(1 + sin θ), so

θ = sin^{−1} \frac{x − 3}{3}, dx = 3 cos θ dθ and \sqrt{6x − x^{2}} = \sqrt{3^{2} − (x − 3)^{2}} =

3 cos θ. This gives

∫ \frac{x^{2} dx}{\sqrt{9 − (x − 3)^{2}}} = ∫ \frac{9(1 + sin θ)^{2} · 3 cos θ dθ}{3 cos θ}
= 9 ∫ (1 + sin θ)^{2} dθ
= 9 ∫ (1 + 2 sin θ + sin^{2} θ) dθ
= 9 ∫ (\frac{3}{2} + 2 sin θ − \frac{1}{2} cos(2θ)) dθ
= 9(\frac{3}{2}θ − 2 cos θ − \frac{1}{4} sin(2θ)) + C
= \frac{27}{2}θ − 18 cos θ − \frac{9}{2} sin θ cos θ + C

Using the formulas 3 sin θ = x − 3 and 3 cos θ = \sqrt{6x − x^{2}} and combining like terms, this becomes

\begin{matrix}\frac{27}{2} sin^{−1}(\frac{x − 3}{3}) − 6\sqrt{6x − x^{2}} − \frac{1}{2}(x − 3)\sqrt{6x − x^{2}} + C \\ = \frac{27}{2} sin^{−1}(\frac{x − 3}{3}) − \frac{1}{2}(x + 9)\sqrt{6x − x^{2}} + C. \end{matrix}

• §10.6 Find the following integrals:

#5. ∫ \frac{2x + 21}{x^{2} − 7x} dx.

Solution: Since x^{2} − 7x = x(x − 7) we write

\frac{2x + 21}{x^{2} − 7x} = \frac{A}{x} + \frac{B}{x − 7},

and find A and B by solving 2x + 21 = A(x − 7) + Bx. Setting x = 0 gives 21 = −7A, so A = −3, and setting x = 7 gives 35 = 7B, so B = 5. Therefore

∫ \frac{2x + 21}{x^{2} − 7x} dx = ∫ \frac{−3 dx}{x} + ∫ \frac{5 dx}{x − 7} = −3 ln |x| + 5 ln |x − 7| + C.

#12. ∫ \frac{4x^{2} + 2x + 4}{x^{3} + 4x} dx.

Solution: Since x^{3} + 4x = x(x^{2} + 4), we write

\frac{4x^{2} + 2x + 4}{x^{3} + 4x} = \frac{Ax + B}{x^{2} + 4} + \frac{C}{x}

and find A, B and C by solving:

4x^{2} + 2x + 4 = (Ax + B)x + C(x^{2} + 4)
= Ax^{2} + Bx + Cx^{2} + 4C = (A + C)x^{2} + Bx + 4C.

Equating coefficients of 1, x and x^{2} gives

A + C = 4; B = 2; 4C = 4,

so C = 1, B = 2 and A = 3. Therefore

∫ \frac{4x^{2} + 2x + 4}{x^{3} + 4x} dx = ∫ \frac{3x dx}{x^{2} + 4} + ∫ \frac{2 dx}{x^{2} + 4} + ∫ \frac{dx}{x}
= \frac{3}{2} ln(x^{2} + 4) + tan^{−1} \frac{x}{2} + ln |x| + C,

where the first integral was found using the substitution u = x^{2} + 4.

#14. ∫ \frac{x^{4} dx}{x^{2} + 4}.

Solution: Long division of x^{4} by x^{2} + 4 gives a quotient of x^{2} − 4 with remainder 16, so

∫ \frac{x^{4} dx}{x^{2} + 4} = ∫ (x^{2} − 4 + \frac{16}{x^{2} + 4}) dx = \frac{1}{3}x^{3} − 4x + 8 tan^{−1} \frac{x}{2} + C.

#24. ∫ \frac{1}{1 + e^{x}} dx.

Solution: First make the substitution u = e^{x}, du = e^{x} dx, so

∫ \frac{1}{1 + e^{x}} dx = ∫ \frac{e^{x} dx}{e^{2x} + e^{x}} = ∫ \frac{du}{u^{2} + u}.

Now u^{2} + u = u(u + 1), so write

\frac{1}{u^{2} + u} = \frac{A}{u} + \frac{B}{u + 1},

and find A and B by solving 1 = A(u + 1) + Bu. Setting u = 0 gives 1 = A, and setting u = −1 gives 1 = −B, so B = −1. Therefore

∫ \frac{du}{u^{2} + u} = ∫ \frac{du}{u} − ∫ \frac{du}{u + 1}
= ln |u| − ln |u + 1| + C
= ln(e^{x}) − ln(e^{x} + 1) + C
= x − ln(e^{x} + 1) + C.

§10.7 Find the following integrals by the method of integration by parts:

#4. ∫ xe^{ax} dx.

Solution: Let u = x, dv = e^{ax} dx. Then du = dx and v = ∫ dv =

\frac{1}{a}e^{ax}, so

∫ xe^{ax} dx = \frac{1}{a}xe^{ax} − \frac{1}{a} ∫ e^{ax} dx = \frac{1}{a}xe^{ax} − \frac{1}{a^{2}} e^{ax} + C.

#8. ∫ sin^{−1} x dx.

Solution: Let u = sin^{−1} x, dv = dx. Then du = (1 − x^{2})^{−1/2} dx and v = x, so

∫ sin^{−1} x dx = x sin^{−1} x − ∫ x(1 − x^{2})^{−1/2} dx.

The substitution w = 1 − x^{2}, dw = −2x dx gives

∫ x(1−x^{2})^{−1/2} dx = −\frac{1}{2} ∫ w^{−1/2}dw = −w^{1/2}+C = −\sqrt{1 − x^{2}}+C.

Therefore

∫ sin^{−1} x dx = x sin^{−1} x + \sqrt{1 − x^{2}} + C.

#13. ∫ x sec^{2} x dx.

Solution: Let u = x, dv = sec^{2} x dx. Then du = dx and v =

∫ dv = tan x, so

∫ x sec^{2} x dx = x tan x − ∫ tan x dx = x tan x − ln | sec x| + C.

#18. ∫ (ln x)^{2} dx.

Solution: Let u = (ln x)^{2}, dv = dx. Then du = \frac{2 ln x}{x}dx and v = x, so

∫ (ln x)^{2} dx = x(ln x)^{2} − 2 ∫ ln x dx.

To find ∫ ln x dx, we use integration by parts again, with u = ln x, dv = dx, giving

∫ (ln x)^{2} dx = x(ln x)^{2} − 2x ln x + 2x + C.

• §10.7 #24.

The region under the curve y = sin x between x = 0 and x = π is revolved about the y-axis. Find the volume of the resulting solid (a) by the shell method; and (b) by the washer method.

Solution: (a) Using shells, dV = 2πx sin x dx, so V = ∫_{0}^{π} 2πx sin x dx.

Let u = x and dv = sin x dx, so du = dx and v = ∫ dv = − cos x. This gives

V = 2π ∫_{0}^{π} x sin x dx = 2π (−x cos x]_{0}^{π} + ∫_{0}^{π} cos x dx)
= 2π (−x cos x + sin x)]_{0}^{π} = 2π^{2}.

(b) The horizontal cross-sections are washers with inner radius x = sin^{−1} y and outer radius x = π − sin^{−1} y. Therefore

dV = π[(π − sin^{−1} y)^{2} − (sin^{−1} y)^{2}] dy = π(π^{2} − 2π sin^{−1} y) dy,

which gives

V = ∫_{0}^{1}(π^{3} − 2π^{2} sin^{−1} y) dy = π^{3} − 2π^{2} ∫_{0}^{1} sin^{−1} y dy.

To evaluate the last integral, let u = sin^{−1} y, dv = dy, so du = (1 − y^{2})^{−1/2}dy and v = y. This gives

∫_{0}^{1} sin^{−1} y dy = y sin^{−1} y]_{0}^{1} − ∫_{0}^{1} y(1 − y^{2})^{−1/2} dy
= \frac{π}{2} + ∫_{1}^{0} \frac{1}{2}w^{−1/2} dw
= \frac{π}{2} + (w^{1/2})]_{1}^{0}
= \frac{π}{2} − 1

(we substituted w = 1 − y^{2}). Therefore

V = π^{3} − 2π^{2}(\frac{π}{2} − 1) = 2π^{2}.

Alternatively, we can convert this integrand dV = π[(π − x)^{2} − x^{2}] dy entirely to x. Since y = sin x then dy = cos x dx. Thus our integral is

∫_{0}^{π/2} π(π^{2} − 2πx) sin x dx

If you check, this gives the same answer.