Integral of $16/(1-\cos8x)$

Question

Can someone please help me with this question:

$$ \int \ \frac{16}{1-\cos8x} \ \ dx \ \ . $$

I tried substitution by letting $u=1-\cos8x$, it got messy after the substitution. I used the identities rules, it still got messy. I don't know what else to do.

Answer 1

The most direct way is probably to multiply and divide by $1+\cos8x$. Then you get $$ 16\,\int\,\frac{1+\cos8x}{1-\cos^28x}\,dx=16\,\int\,\frac{1}{1-\cos^28x}\,dx+16\,\int\,\frac{\cos8x}{1-\cos^28x}\,dx\\ =16\,\int\,\frac{1}{\sin^28x}\,dx+16\,\int\,\frac{\cos8x}{\sin^28x}\,dx, $$ both easy.

Answer 2

Letting $\theta=8x$, the integral becomes

$$\int\frac{16\,dx}{1-\cos{8x}}=2\int\frac{d\theta}{1-\cos{\theta}}.$$

The integral $\int\frac{d\theta}{1-\cos{\theta}}$ can be solved using the world's sneakiest substitution: $t=\tan{\frac{\theta}{2}}$. Under this substitution,

$$\int\frac{d\theta}{1-\cos{\theta}}=\int\frac{dt}{t^2}=-\frac{1}{t}+constant=-\cot{\frac{\theta}{2}}+constant.$$

Altogether,

$$\int\frac{16\,dx}{1-\cos{8x}}=-2\cot{4x}.$$

Answer 3

First let $u = 8x$ and then $du = 8dx.$ So your integral becomes

\begin{align} \int\frac{2}{1 - \cos u} \cdot \frac{1 + \cos u}{ 1 + \cos u} du &= \int \frac{2(1 + \cos u)}{1 - \cos^2 u} du \\ &= \int \frac{ 2(1 + \cos u)}{\sin^2 u} du\\ &= \int 2(\csc^2 u + \cot u \csc u) du \\ &=-2\cot u - 2\csc u + C\\ &=-2(\cot 8x + \csc 8x) + C \end{align}

Answer 4

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} \color{#00f}{\large\int{16 \over 1 - \cos\pars{8x}}\,\dd x}&= 16\int{\dd x\over 2\sin^{2}\pars{4x}} =8\int\csc^{2}\pars{4x}\,\dd x =8\bracks{-\,{\cot\pars{4x} \over 4}} \\[3mm]&=\color{#00f}{\large-2\cot\pars{4x}} + \mbox{a constant} \end{align}