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Find the first four nonzero terms of the power series solution of y′ = x + y2

by RoRi

Consider the differential equation

    \[ y' = x + y^2 \]

with initial conditions y = 0 when x = 0. Assume this differential equation has a power-series solution and compute the first four nonzero terms of the expansion.

Let

    \[ f(x) = \sum_{n=0}^{\infty} a_n x^n.\]

be the power-series solution of the differential equation. Then we must have

    \begin{align*} && \sum_{n=1}^{\infty} na_n x^{n-1} &= x + \left( \sum_{n=0}^{\infty} a_n x^n \right)^2 \\[9pt] \implies && \sum_{n=1}^{\infty} na_n x^{n-1} &= x + \sum_{n=0}^{\infty} \left( \sum_{k=0}^n a_k a_{n-k} \right) x^n \right) \\[9pt] \implies && a_1 + 2a_2 x + 3a_3 x^2 + \cdots &= a_0^2 + (2a_0 a_1 + 1)x + (a_1^2 + 2a_0a_2)x^2 + \cdots. \end{align*}

From the initial conditions we know a_0 = 0. Then, equating like powers of x we can solve for the first four nonzero terms in the power series expansion:

    \begin{align*} a_0 &= 0 \\[9pt] a_1 &= a_0^2 & \implies && a_1 &= 0 \\[9pt] 2a_2 &= 2a_0 a_1 + 1 & \implies && a_2 & = \frac{1}{2} \\[9pt] 3a_3 &= a_1^2 + 2a_0 a_2 & \implies && a_3 &= 0 \\[9pt] 4a_4 &= 2(a_0 a_3 + a_1 a_2) & \implies && a_4 &= 0 \\[9pt] 5a_5 &= a_2^2 + 2(a_0 a_4 + a_1 a_3) & \implies && a_5 &= \frac{1}{20} \\[9pt] 6a_6 &= 2(a_0 a_5 + a_1 a_4 + a_2 a_3) & \implies && a_6 &= 0 \\[9pt] 7a_7 &= a_3^2 + 2(a_0 a_6 + a_1 a_5 + a_2 a_4) & \implies && a_7 &= 0 \\[9pt] 8a_8 &= 2(a_0 a_7 + a_1 a_6 + a_2 a_5 + a_3 a_4) & \implies && a_8 &= \frac{1}{160} \\[9pt] 9a_9 &= a_4^2 + 2(a_0 a_8 + a_1 a_7 + a_2 a_6 + a_3 a_5) & \implies && a_9 &= 0 \\[9pt] 10a_{10} &= 2(a_0 a_9 + a_1 a_8 + a_2 a_7 + a_3 a_6 + a_4 a_5) & \implies && a_{10} &= 0 \\[9pt] 11a_{11} &= a_5^2 + 2(a_0 a_{10} + a_1 a_9 + a_2 a_8 + a_3 a_7 + a_4 a_6) & \implies && a_{11} &= \frac{1}{8800} \end{align*}

(Note: I think the solution in the back of Apostol is wrong on this. Apostol has a_5 = \frac{1}{12}, a_8 = \frac{1}{060}, and a_{11} = \frac{7}{8800}. I’m going to mark this as errata until someone convinces me Apostol is actually correct.)

Therefore, we have

    \[ y = \frac{1}{2} x^2 + \frac{1}{20} x^5 + \frac{1}{160} x^8 + \frac{1}{8800} x^{11} + \cdots. \]

Find the first four nonzero terms of the power series solution of y′ = 1 + xy2

by RoRi

Consider the differential equation

    \[ y' = 1 + xy^2 \]

with initial conditions y = 0 when x = 0. Assume this differential equation has a power-series solution and compute the first four nonzero terms of the expansion.

Let

    \[ f(x) = \sum_{n=0}^{\infty} a_n x^n \]

be the power series solution of the differential equation. Then we must have

    \begin{align*} && \sum_{n=1}^{\infty} na_n x^{n-1} &= 1 + x \left( \sum_{n=0}^{\infty} a_n x^n \right)^2 \\[9pt] \implies && \sum_{n=1}^{\infty} na_n x^{n-1} &= 1 + x \left( \sum_{n=0}^{\infty} \left( \sum_{k=0}^n a_k a_{n-k} \right) x^n \right) \\[9pt] \implies && a_1 + 2a_2 x + 3a_3 x^2 + \cdots & = 1 + a_0^2 x + (2a_0 a_1)x^2 + (a_1^2 + 2a_0a_2)x^3 + \cdots. \end{align*}

From the initial condition y =0 when x = 0 we know a_0 = 0. Therefore, equating like powers of x we have

    \begin{align*} a_1 &= 1 \\[9pt] 2a_2 &= a_0^2 & \implies && a_2 &= 0 \\[9pt] 3a_3 &= 2a_0 a_1 & \implies && a_3 &= 0 \\[9pt] 4a_4 &= a_1^2 + 2a_0 a_2 & \implies && a_4 &= \frac{1}{4} \\[9pt] 5a_5 &= 2(a_0 a_3 + a_1 a_2) & \implies && a_5 &= 0 \\[9pt] 6a_6 &= a_2^2 + 2(a_0 a_4 + a_1 a_3) & \implies && a_6 &= 0 \\[9pt] 7a_7 &= 2(a_0 a_5 + a_1 a_4 + a_2 a_3) & \implies && a_7 &= \frac{1}{14} \\[9pt] 8a_8 &= a_3^2 + 2(a_0 a_6 + a_1 a_5 + a_2 a_4) & \implies && a_8 &= 0 \\[9pt] 9a_9 &= 2(a_0 a_7 + a_1 a_6 + a_2 a_5 + a_3 a_4) & \implies && a_9 &= 0 \\[9pt] 10a_{10} &= a_4^2 + 2(a_0 a_8 + a_1 a_7 + a_2 a_6 + a_3 a_5) & \implies && a_{10} &= \frac{23}{112}. \end{align*}

(Note: The book gives the value a_{10} = \frac{23}{1120}. I think the answer we have above is correct. I’m marking this as errata unless someone convinces me that Apostol is correct.)

Therefore, we have

    \[ f(x) = x + \frac{1}{4} x^4 + \frac{1}{14} x^7 + \frac{23}{112} x^{10} + \cdots. \]

Determine all real numbers satisfying given relations

by RoRi

Determine all values for the real numbers x and y such that the following equations hold.

  1. x + iy = x - iy.
  2. x + iy = |x + iy|.
  3. |x+iy| = |x-iy|.
  4. (x+iy)^2 = (x-iy)^2.
  5. \frac{x+iy}{x-iy} = x - iy.
  6. \sum_{k=0}^{100} i^k = x + iy.
  1. The equation

        \[ x+iy = x-iy \quad \implies \quad 2iy = 0 \quad \implies \quad y = 0. \]

    The value of x is arbitrary.

  2. Using the formula for the absolute value of a complex number we have

        \begin{align*} x+iy = |x+iy| && \implies && x+iy = \sqrt{x^2 + y^2} \\ && \implies && x = \sqrt{x^2+y^2} \text{ and } y = 0. \end{align*}

    Since y = 0 the equation x = \sqrt{x^2+y^2} implies x = \sqrt{x^2} which implies x \geq 0. Therefore, this equation is satisfied by

        \[ x \geq 0, \qquad y = 0. \]

    (Note: The answer Apostol gives says x > 0, but I think x = 0 works as well.

  3. Again, using the formula for the absolute value of a complex number we have

        \begin{align*} |x + iy| = |x-iy|&& \implies && \sqrt{x^2 + y^2} &= \sqrt{x^2 + (-y)^2} \\ && \implies && \sqrt{x^2+y^2} &= \sqrt{x^2 + y^2}. \end{align*}

    This holds for all real x and y.

  4. We compute as follows,

        \begin{align*} (x+iy)^2 = (x-iy)^2 && \implies && x^2 - y^2 + (2xy)i &= x^2 - y^2 - (2xy)i \\ && \implies && (4xy)i = 0. \end{align*}

    Hence, we must have either x = 0 and y is arbitrary or x arbitrary and y =0.

  5. We compute,

        \begin{align*} \frac{x+iy}{x-iy} = x-iy && \implies && x+iy &= (x-iy)^2 \\ && \implies && x+iy &= x^2 - y^2 - (2xy)i. \end{align*}

    This gives us two equations (since the real parts and imaginary parts must be equal),

        \[ x = x^2 - y^2, \qquad y = -2xy. \]

    If y \neq 0 then from the second equation we have

        \[ x = -\frac{1}{2} \quad \implies \quad -\frac{1}{2} = \left( -\frac{1}{2} \right)^2 - y^2 \quad \implies \quad y = \frac{\sqrt{3}}{2}. \]

    If y = 0 then we have x = x^2 so x = 0 or x = 1. But, x = 0 is not impossible since then \frac{x+iy}{x-iy} is undefined. Therefore we have two possibilities

        \[ x = 1, \ y = 0 \qquad \text{or} \qquad x = -\frac{1}{2}, \ y = \frac{sqrt{3}}{2}. \]

    (Note: Apostol only gives the first of these solutions. We can check by a direct substitution that the second solution also works.)

  6. Here we note that

        \[ i^k = \begin{cases} 1 &\text{if } k \equiv 0 \mod 4 \\ i & \text{if } k \equiv 1 \mod 4 \\ -1 & \text{if } k \equiv 2 \mod 4 \\ -i & \text{if } k \equiv 3 \mod 4. \end{cases} \]

    Therefore, we have

        \begin{align*} \sum_{k=0}^{100} i^k &= \sum_{k=0}^{25} i^{4k} + \sum_{k=0}^{24} i^{4k+1} + \sum_{k=0}^{24} i^{4k+2} + \sum_{k=0}^{24} i^{4k+3} \\ &= \sum_{k=0}^{25} 1 + \sum_{k=0}^{24} i + \sum_{k=0}^{24} (-1) + \sum_{k=0}^{24} (-i) \\ &= 26 + 25i - 25 - 25i \\ &= 1 \end{align*}

    Therefore, from the given equation we have

        \[ x + iy =1 \quad \implies \quad x = 1, \ \ y = 0. \]

Compute the limit of the given function

by RoRi

Evaluate the limit.

    \[ \lim_{x \to 0} \left( \frac{1}{\log \left( x + \sqrt{1+x^2} \right)} - \frac{1}{\log(1+x)} \right). \]

( Note: There’s a small typo in Apostol which puts a parenthesis in the wrong place. The statement above is what I assume is meant since this evaluates to the answer given in the back of the book.)

To do this we’ll need to get the expression in the limit into the indeterminate form 0/0 and then apply L’Hopital’s rule twice. (Applying L’Hopital’s is going to be a challenge since the derivatives are going to get quite messy before we get anywhere.) We start by putting things over a common denominator,

    \[ \lim_{x \to 0} \left( \frac{1}{\log \left( x + \sqrt{1+x^2} \right)} - \frac{1}{\log (1+x)} \right) &= \lim_{x \to 0} \left( \frac{ \log(1+x) - \log \left( x + \sqrt{1+x^2} \right)}{\log \left( x + \sqrt{1+x^2} \right) \log (1+x)} \right). \]

Now we want to apply L’Hopital’s. First, we take the derivative of the numerator,

    \begin{align*} D \left( \log (1+x) + \log \left( x + \sqrt{1+x^2} \right) \right) &= \frac{1}{1+x} + \frac{1}{x+\sqrt{1+x^2}} \cdot \left( 1 + \frac{x}{\sqrt{1+x^2}} \right) \\[9pt] &= \frac{1}{1+x} + \frac{ 1 + \frac{x}{\sqrt{1+x^2}}}{x + \sqrt{1+x^2}} \\[9pt] &= \frac{1}{1+x} + \frac{\sqrt{1+x^2} + x}{\sqrt{1+x^2} \left( x + \sqrt{1+x^2} \right)} \\[9pt] &= \frac{1}{1+x} + \frac{1}{\sqrt{1+x^2}}. \end{align*}

Next, we take the derivative of the denominator,

    \begin{align*} D \left( \log \left( 1 + \sqrt{1+x^2} \right) \log (1+x) \right) &= \frac{1 + \frac{x}{\sqrt{1+x^2}}}{1+\sqrt{1+x^2}} \cdot \log(1+x) + \frac{1}{1+x} \cdot \log \left( 1 + \sqrt{1+x^2} \right) \\[9pt] &= \frac{ (1+x) \log (1+x) + \sqrt{1+x^2} \log \left( x + \sqrt{1+x^2} \right)}{(1+x)\sqrt{1+x^2}}. \end{align*}

So, now we use proceed with L’Hopital’s. (Keep in mind that L’Hopital’s is only valid if, once we get to the end, the limits of the derivatives we have taken actually exist, so right now we’re taking derivatives and hoping that the limits exist… once we have established that they do, then the whole process was valid.)

    \begin{align*} \lim_{x \to 0} \left( \frac{ \log(1+x) - \log \left( x + \sqrt{1+x^2} \right)}{\log \left( x + \sqrt{1+x^2} \right) \log (1+x)} \right) &= \lim_{x \to 0} \frac{ \frac{1}{1+x} + \frac{1}{\sqrt{1+x^2}}}{ \frac{ (1+x) \log (1+x) + \sqrt{1+x^2} \log \left( x + \sqrt{1+x^2} \right)}{(1+x)\sqrt{1+x^2}} } \\[9pt] &= \lim_{x \to 0}\frac{ (1+x)\sqrt{1+x^2} \left( \frac{1}{1+x} + \frac{1}{\sqrt{1+x^2}} \right)}{(1+x)\log(1+x) + \sqrt{1+x^2} \log \left(x + \sqrt{1+x^2} \right)} \\[9pt] &= \lim_{x \to 0} \frac{ \sqrt{1+x^2} - (1+x)}{(1+x)\log(1+x) + \sqrt{1+x^2} \log \left(x + \sqrt{1+x^2} \right)}. \end{align*}

So, we again have the indeterminate form 0/0, and again we’ll try to apply L’Hopital’s. The derivative of the numerator is

    \begin{align*} D \left( \sqrt{1+x^2} - (1+x) \right) &= \frac{x}{\sqrt{1+x^2}} -1 \\[9pt] &= \frac{x - \sqrt{1+x^2}}{\sqrt{1+x^2}}. \end{align*}

The derivative of the denominator is

    \begin{align*} D &\left( (1+x)\log(1+x) + \sqrt{1+x^2} \log \left(x + \sqrt{1+x^2} \right) \right) \\[9pt] &= \log(1+x) + (1+x) \frac{1}{1+x} + \frac{x}{\sqrt{1+x^2}} \log \left( x + \sqrt{1+x^2} \right) + \sqrt{1+x^2} \frac{1 + \frac{x}{\sqrt{1+x^2}}}{x + \sqrt{1+x^2}} \\[9pt] &= 2 + \log(1+x) + \frac{x \log \left( x + \sqrt{1+x^2} \right)}{\sqrt{1+x^2}}. \end{align*}

Putting these back into our evaluation of the limit,

    \begin{align*} \lim_{x \to 0} \frac{ \sqrt{1+x^2} - (1+x)}{(1+x)\log(1+x) + \sqrt{1+x^2} \log \left(x + \sqrt{1+x^2} \right)} &= \lim_{x \to 0} \frac{ \frac{x - \sqrt{1+x^2}}{\sqrt{1+x^2}} }{2 + \log(1+x) + \frac{x \log \left( x + \sqrt{1+x^2} \right)}{\sqrt{1+x^2}}}. \end{align*}

But now, we have a quotient of continuous functions with non-zero denominator when x = 0 so the limit is the value of the function at x = 0. Evaluating we then have

    \[ \lim_{x \to 0} \left( \frac{1}{\log \left( x + \sqrt{1+x^2} \right)} - \frac{1}{\log(1+x)} \right) = -\frac{1}{2}.\]

Evaluate the integral of ((x-a)(b-x))-1/2

by RoRi

Evaluate the following integral for b \neq a.

    \[ \int \frac{dx}{\sqrt{(x-a)(b-x)}}. \]

(Note: Again, there is an error in the answers in the back of the book for this problem and the previous one. The answers listed for problems #46 and #47 should be swapped.)

Following the hint, we make the substitution

    \begin{align*} && x-a &= (b-a) \sin^2 u \\ \implies && b-x &= (b-a) (1- \sin^2 u) \\ \implies && u &= \arcsin \sqrt{\frac{x-a}{b-a}} \\ \implies && dx &= 2(b-a) \sin u \cos u \, du. \end{align*}

Therefore, we can rewrite this integral in terms of u,

    \begin{align*} \int \frac{dx}{\sqrt{(x-a)(b-x)}} &= \int \frac{2(b-a) \sin u \cos u \, du}{\sqrt{(b-a)^2 \sin^2 u (1 - \sin^2 u)}} \\ &= \frac{2(b-a)}{|b-a|} \int \frac{\cos u \, du}{\sqrt{1-\sin^2 u}} \\ &= \frac{2(b-a)}{|b-a|} \int du \\ &= \frac{2(b-a)}{|b-a} u + C. \end{align*}

Substituting back in for u we then have

    \begin{align*} \int \frac{dx}{\sqrt{(x-a)(b-x)}} &= \frac{2(b-a)}{|b-a|} u + C \\ &= \frac{2(b-a)}{|b-a|} \arcsin \sqrt{\frac{x-a}{b-a}} + C. \end{align*}

Evaluate the integral of ((x-a)(b-x))1/2

by RoRi

Evaluate the following integral for b \neq a.

    \[ \int \sqrt{(x-a)(b-x)} \, dx. \]

(Note: This is a pretty involved problem the way I’ve done it. Maybe there’s a better way? Let me know if you have one. Also, there is an error in the answer in the book on this problem and the next one. The answers given in the book are swapped, so the answer listed for this problem #46 is actually the answer for #47 and vice-versa.)

There are some integrals we’ll want to use to carry out the evaluation of the above integral. First, from previous exercises here and here (Section 5.10, Exercises #7 and #10(b)) we know

    \begin{align*} \int \sin^2 x \, dx &= \frac{1}{2} x - \frac{1}{4} \sin (2x) \\ \int \sin^4 x \, dx &= \frac{3}{8} x - \frac{1}{4} \sin (2x) + \frac{1}{32} \sin (4x). \end{align*}

Therefore, (we’ll want this later), we have

    \begin{align*} \int (\sin^2 x - \sin^4 x ) \, dx &= \left( \frac{1}{2} x - \frac{1}{4} \sin (2x) \right) - \left( \frac{3}{8} x - \frac{1}{4} \sin (2x) + \frac{1}{32} \sin (4x) \right) \\ &= \frac{1}{8} x - \frac{\sin (4x)}{32} + C\\ &= \frac{1}{8} x - \frac{\sin (2x) \cos (2x)}{16} + C\\ &= \frac{1}{8} x - \frac{\sin x \cos x \cos (2x)}{8} + C\\ &= \frac{1}{8} \left( x - \sin x \cos x (1 - 2 \sin^2 x) \right) + C\\ &= \frac{1}{8} \left( x - \sin x \cos x + 2 \sin^3 x \cos x \right) + C. \end{align*}

The other integral that we are going to want to have available is

    \[ \int x^2 \sqrt{1-x^2} \, dx. \]

To evaluate this we’ll use the trig integrals above. First, make the substitution

    \[ \sin t = x \qquad \implies \qquad \cos t \, dt = dx \]

and also gives us

    \[ t = \arcsin x \qquad \text{and} \qquad dt = \frac{dx}{\cos t}. \]

Therefore we have

    \begin{align*} \int x^2 \sqrt{1-x^2} \, dx &= \int \sin^2 t \sqrt{1 - \sin^2 t} \cos t \, dt \\ &= \int \sin^2 t \sqrt{\cos^2 t} \cos t \, dt \\ &= \int \sin^2 t \cos^2 t \, dt \\ &= \int \sin^2 t (1 - \sin^2 t) \, dt \\ &= \int (\sin^2 t - \sin^4 t) \, dt \\ &= \frac{1}{8}\left( t - \sin t \cos t + 2 \sin^3 t \cos t \right) + C. \end{align*}

Then, substituting back in for x (and noting that \sin (\arcsin x) = x and \cos (\arcsin x) = \sqrt{1-x^2}) we have

    \begin{align*} \int x^2 \sqrt{1-x^2} \, dx &= \frac{1}{8} \left( t - \sin t \cos t + 2 \sin^3 t \cos t \right) + C\\ &= \frac{1}{8} \left( \arcsin x - x \sqrt{1-x^2} + 2 x^3 \sqrt{1-x^2} \right) + C \\ &= \frac{1}{8} \arcsin x - \frac{1}{8} x \sqrt{1-x^2} (1 - 2x^2) + C. \end{align*}

So, now that we have those, we can turn our attention to the integral in the question. For this integral we want to make the substitution

    \[ t = \frac{x-a}{b-a} \quad \implies \quad dt = \frac{dx}{b-a}. \]

which implies

    \[ x = t(b-a) + a \quad \text{and} \quad dx = (b-a) \, dt. \]

Therefore we have

    \begin{align*} \int \sqrt{(x-a)(b-x)} \, dx &= \int \sqrt{(t(b-a)+a - a)(b - t(b-a) - a)} \, (b-a)dt \\ &= (b-a) \int \sqrt{t(b-a)(b-a)(1-t)} \, dt \\ &= (b-a) |b-a| \int \sqrt{t} \sqrt{1-t} \, dt. \end{align*}

Now, we want to make the substitution

    \[ u = \sqrt{t} \quad \implies \quad du = \frac{1}{2\sqrt{t}} \, dt, \]

and implies

    \[ t = u^2 \quad \text{and} \quad dt = 2u \, du. \]

Therefore,

    \begin{align*} \int \sqrt{(x-a)(b-x)} \, dx &= (b-a)|b-a| \int \sqrt{t} \sqrt{1-t} \, dt \\ &= (b-a)|b-a| \int u \sqrt{1-u^2} 2u \, du \\ &= 2 (b-a)|b-a| \int u^2 \sqrt{1-u^2} \, du. \end{align*}

Now, we can use the work we did above in the evaluation of this integral,

    \begin{align*} \int \sqrt{(x-a)(b-x)} \, dx &= 2 (b-a)|b-a| \int u^2 \sqrt{1-u^2} \, du \\ &= 2(b-a)|b-a| \left( \frac{1}{8} \arcsin u - \frac{1}{8} u \sqrt{1-u^2} (1-2u^2) \right) + C \\ &= \frac{1}{4}|b-a|(b-a) \left(\arcsin u + u \sqrt{1-u^2} (2u^2 - 1)\right) + C. \end{align*}

Finally, we have to unwind our substitutions to get back to a function of x. We have

    \[ u = \sqrt{t} = \sqrt{ \frac{x-a}{b-a}}. \]

Therefore,

    \begin{align*} \int \sqrt{(x-a)(b-x)} \, dx &= \frac{1}{4}|b-a|(b-a) \left(\arcsin u + \frac{1}{4} u \sqrt{1-u^2} (2u^2 - 1)\right) + C\\[10pt] &= \frac{1}{4} |b-a|(b-a) \arcsin \left( \sqrt{\frac{x-a}{b-a}} \right) + \\ &\qquad \frac{1}{4} |b-a|(b-a) \left( \sqrt{\frac{x-a}{b-a}} \right) \left(\sqrt{ 1 - \frac{x-a}{b-a} } \right) \left( 2 \frac{x-a}{b-a} - 1 \right) + C \\[10pt] &= \frac{1}{4} |b-a|(b-a) \arcsin \left( \sqrt{\frac{x-a}{b-a}} \right) + \\ &\qquad \frac{1}{4} |b-a|(b-a) \sqrt{ \frac{(x-a)(b-a) - (x-a)^2}{(b-a)^2} } \cdot \frac{2x-2a-(b-a)}{b-a}+ C \\[10pt] &= \frac{1}{4}|b-a|(b-a) \arcsin \sqrt{ \frac{x-a}{b-a}} \\ & \qquad + \frac{1}{4} \sqrt{x-a} \sqrt{b-a - (x-a)} \cdot (2x - b - a) + C \\[10pt] &= \frac{1}{4}|b-a|(b-a) \arcsin \sqrt{\frac{x-a}{b-a}} + \frac{1}{4} \sqrt{(x-a)(b-x)} (2x - (a+b)) + C. \end{align*}

This completes our evaluation of the integral.

Find indefinite integrals of powers of sin

by RoRi

From the previous two exercises (here and here) we know

    \[ \int \sin^2 x \, dx = \frac{1}{2}x - \frac{1}{4} \sin (2x), \]

and

    \[ \int \sin^n x \, dx = - \frac{\sin^{n-1} x \cos x}{n} + \frac{n-1}{n} \int \sin^{n-2} x \, dx. \]

Use these results to establish the following formulas:

  1. \displaystyle{ \int \sin^3 x \, dx = -\frac{3}{4} \cos x + \frac{1}{12} \cos (3x)}.
  2. \displaystyle{ \int \sin^4 x \, dx = \frac{3}{8} x - \frac{1}{4} \sin (2x) + \frac{1}{32} \sin (4x)}.
  3. \displaystyle{ \int \sin^5 x \, dx = -\frac{5}{8} \cos x + \frac{5}{48} \cos (3x) - \frac{1}{80} \cos (5x)}.

    4. ( Note: There is an error in my edition of the book for part (c). It requests we prove \int \sin^5 x \, dx = -\frac{5}{8} x + \frac{5}{48} \cos (3x) - \frac{1}{80} \cos(5x), omitting the \cos x in the first term of the result. This is just an error in the book since the formula as stated is false.)

  1. Using the second formula, and the triple angle identity for cosine (\cos (3x) = 4 \cos^3 x - 3 \cos x which implies \cos^3 x = \frac{1}{4} \cos (3x) + \frac{3}{4} \cos x, derived in this exercise), we have

        \begin{align*} \int \sin^3 x \, dx &= - \frac{\sin^2 x \cos x}{3} + \frac{2}{3} \int \sin x \, dx \\[9pt] &= -\frac{1}{3} \left( \sin^2 x \cos x + 2 \cos x ) \\[9pt] &= -\frac{1}{3} \left( (1 - \cos^2 x) \cos x + 2 \cos x ) \\[9pt] &= -\frac{1}{3} \cos x (3 - \cos^2 x) \\[9pt] &= - \cos x + \frac{1}{3} \cos^3 x \\[9pt] &= - \cos x + \frac{1}{3} \left(\frac{1}{4} \cos (3x) + \frac{3}{4} \cos x \right) \\[9pt] &= -\frac{3}{4} \cos x + \frac{1}{12} \cos (3x). \qquad \blacksquare \end{align*}

  2. Using the second formula above and then the first we have (and also recalling the trig identity \cos (2x) = 1 - 2 \sin^2 x which implies \sin^2 x = \frac{1}{2} - \frac{1}{2} \cos (2x)),

        \begin{align*} \int \sin^4 x \, dx &= - \frac{\sin^3 x \cos x}{4} + \frac{3}{4} \int \sin^2 x \, dx \\[9pt] &= -\frac{1}{4} (\sin^3 x \cos x) + \frac{3}{4} \left( \frac{1}{2} x - \frac{1}{4} \sin (2x) \right) \\[9pt] &= -\frac{1}{4} (\sin^2 x \sin x \cos x) + \frac{3}{8} x - \frac{3}{16} \sin (2x) \\[9pt] &= -\frac{1}{8} (\sin^2 x \sin (2x)) + \frac{3}{8} x - \frac{3}{16} \sin (2x) \\[9pt] &= -\frac{1}{16} (\sin (2x) - \sin (2x) \cos (2x)) + \frac{3}{8} x - \frac{3}{16} \sin (2x) \\[9pt] &= \frac{3}{8} - \frac{1}{4} \sin (2x) + \frac{1}{32} \sin (4x). \qquad \blacksquare \end{align*}

  3. Using the second formula above and then part (a) we have

        \begin{align*} \int \sin^5 x \, dx &= -\frac{\sin^4 x \cos x}{5} + \frac{4}{5} \int \sin^3 x \, dx \\[9pt] &= -\frac{1}{5} \left( (1-\cos^2 x)(1-\cos^2 x) \cos x) + \frac{4}{5} \left(-\frac{3}{4} \cos x + \frac{1}{12} \cos (3x) \right) \\[9pt] &= -\frac{1}{5} (\cos x - 2 \cos^3 x + \cos^5 x) + \frac{1}{16} (10 \cos x + 5 \cos (3x) + \cos (5x)) \\[9pt] & \qquad \qquad - \frac{3}{5} \cos x + \frac{1}{15} \cos (3x) \\[9pt] &= -\frac{1}{5} \left( \frac{1}{8} \cos x - \frac{3}{16} \cos (3x) + \frac{1}{16} \cos (5x) \right) - \frac{3}{5}\cos x + \frac{1}{15} \cos (3x) \\[9pt] &= -\frac{5}{8} \cos x + \frac{5}{48} \cos (3x) - \frac{1}{80} \cos (5x). \qquad \blacksquare \end{align*}

Establish the given limit formula

by RoRi

Prove the formula for the limit:

    \[ \lim_{x \to {\color{red}a}} \frac{\sin x - \sin a}{x - a} = \cos a. \]

Using,

    \[ \lim_{x \to {\color{red}a}} \frac{\sin x}{x} = 1. \]

( Note: I think there is an error in this problem as stated in Apostol. The statement in the book is for the limit as x \to 0. However, the limit as x \to 0 of this is -\frac{\sin a}{a} not \cos a (since it is a quotient of continuous functions and for a \neq 0 the denominator is nonzero at x = 0). I’ve changed the statement of the question to be the limit as x approaches a. This makes the given formula correct, so is probably what was intended.)

Proof. From Theorem 2.3 (g) (Apostol, p. 96) we know

    \[ \sin a - \sin b = 2 \sin \frac{a-b}{2} \cos \frac{a+b}{2}. \]

So, evaluating the limit we have,

    \begin{align*} \lim_{x \to a} \frac{\sin x - \sin a}{x-a} &= \lim_{x \to a} \frac{2 \sin \frac{x-a}{2} \cos \frac{x+a}{2}}{x-a} \\ &= \lim_{x \to a} \left( \frac{\sin \frac{x-a}{2}}{\frac{x-a}{2}} \cos \frac{x+a}{2} \right). \end{align*}

But then,

    \[ \lim_{x \to a} \frac{\sin \frac{x-a}{2}}{\frac{x-a}{2}} = \lim_{y \to 0} \frac{\sin y}{y} = 1.\]

where,

    \[ y = \frac{x-a}{2} \quad \text{ and } \quad y \to 0 \quad \text{ as } \quad x \to a. \]

So,

    \begin{align*} \lim_{x \to a} \frac{\sin \frac{x-a}{2}}{\frac{x-a}{2}} \cos \frac{x+a}{2} &= \lim_{x \to a} \cos \frac{x+a}{2} \\ &= \cos a. \qquad \blacksquare \end{align*}