A calculus of the absurd

11.5.2 Integral of \(\frac {\sin (x)}{\cos (x) + \cos ^3(x)}\)

Example 11.5.2 Find the value of⁸³⁸³ This question came from https://madasmaths.com
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\begin{equation*} \int \frac {\sin (x)}{\cos (x) + \cos ^3(x)} dx \end{equation*}

Solution: We substitute \(u=\sin (x)\). If this seems like a strange thing to do, there are a bunch of good reasons:

• We know that the derivative of \(\cos (x)\) is \(-\sin (x)\), so we can replace \(\sin (x)\) by \(-\frac {du}{dx}\), which we then integrate w.r.t \(x\), so the overall integral is integrated w.r.t \(u\).
• As we can replace \(\cos (x)\) with \(u\) and \(\cos (x)\) with \(u^3\), this simplifies the expression considerably.

Proceeding,

\begin{align*} \int \frac {\sin (x)}{\cos (x) + \cos ^3(x)} dx &= \int \frac {-\frac {du}{dx}}{u + u^3} dx \\ &= - \int \frac {1}{u + u^3} du \\ &= - \int \frac {1}{u(1 + u^2)} du \end{align*}

Which we can split into partial fractions:

\begin{align*} & \frac {1}{u(1 + u^2)} = \frac {A}{u} + \frac {Bu + C}{1+u^2} \\ &\implies 1 = A(1+u^2) + (Bu + C)u \\ &\implies 0 = A + Au^2 + Bu^2 + Cu - 1 \\ &\implies 0 = (A+B)u^2 + Cu + (A - 1) \end{align*}

From where we can come up with some simultaneous equations

\begin{equation*} \begin{cases} A + B = 0 \\ C = 0 \\ A - 1 = 0 \end {cases} \end{equation*}

From this we can deduce that \(A = 1\), \(B = -1\) and \(C=0\). Overall, then, we have

\begin{align*} - \int \frac {1}{u} - \frac {u}{1+u^2} du &= \int \frac {u}{1+u^2} - \frac {1}{u} du \\ &= \frac {1}{2} \ln (1+u^2) - \ln (u) + c \\ &= \ln \left ( \frac {\sqrt {1+u^2}}{u} \right ) + c \\ &= \ln \left ( \sqrt {\frac {1+u^2}{u^2}} \right ) + c \\ &= \frac {1}{2} \ln \left ( \frac {1+u^2}{u^2} \right ) + c \end{align*}

Note that another way to do the last four steps is

\begin{align*} \frac {1}{2} \ln (1+u^2) - \ln (u) &= \frac {1}{2} \left [ \ln (1+u^2) - 2\ln (u) \right ] \\ &= \frac {1}{2} \left [ \ln \left ( \frac {1+u^2}{u^2} \right ) \right ] \end{align*}

which is possibly nicer.