Please help me. I have been stuck on this for ages :(

$$\int \frac{1}{13\cos x+ 12}\,\mathrm{d}x$$

I appreciate any and all help. Thank you.



For the integrals of the form $\displaystyle\frac1{a\sin x+b\cos x+c},$ where $a,b,c$ are arbitrary constants

try setting $\displaystyle\tan\frac x2=t$ and use Weierstrass substitution

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  • 1
    $\begingroup$ This was an excellent hint. Solving the problem was only secondary in comparison to learning a new technique for integration. Thank you very much! $\endgroup$ – A is for Ambition Jun 15 '14 at 12:01

$\bf{My\; Solution::}$ Given $$\displaystyle \int\frac{1}{13\cos x+12}dx = \int\frac{1}{13(1+\cos x)-1}dx$$

Now Using $$\displaystyle 1+\cos x = 2\cos^2 \frac{x}{2}\;,$$ we get

$$\displaystyle \int\frac{1}{26\cos^2 \frac{x}{2}-1}dx$$ Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\displaystyle \cos^2 \frac{x}{2}$

$$\displaystyle \int\frac{\sec^2 \frac{x}{2}}{26-1-\tan^2 \frac{x}{2}}dx = \int\frac{\sec^2 \frac{x}{2}}{5^2-\tan^2\frac{x}{2}}dx$$

Now Let $\displaystyle \tan \frac{x}{2} = t\;,$ Then $\displaystyle \sec^2\frac{x}{2}dx = 2dt$

So Integral is $$\displaystyle 2\int\frac{1}{5^2-t^2}dt = \frac{1}{10}\ln \left|\frac{5+t}{5-t}\right|+\mathbb{C}$$

Where $\displaystyle t = \cos^2 \frac{x}{2}$

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  • $\begingroup$ Was going to add this as a separate answer:) Good job. $\endgroup$ – lab bhattacharjee Jun 15 '14 at 11:39
  • $\begingroup$ Soory lab bhattacharjee actually I did not seen yrs Hint. Thanks $\endgroup$ – juantheron Jun 15 '14 at 11:43
  • $\begingroup$ Nothing to be sorry for. $\endgroup$ – lab bhattacharjee Jun 15 '14 at 11:44

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} &\color{#66f}{\large\int{1 \over 13\cos\pars{x} + 12}\,\dd x} =\int{1 \over 26\cos^{2}\pars{x/2} - 1}\,\dd x =\int{\sec^{2}\pars{x/2} \over 26 - \sec^{2}\pars{x/2}}\,\dd x \\[3mm]&=\ \overbrace{\int{\sec^{2}\pars{x/2} \over 25 - \tan^{2}\pars{x/2}}\,\dd x} ^{\ds{\tan\pars{x/2} \equiv t}}\ =\ 2\int{\dd t \over 25 - t^{2}} ={1 \over 5}\int\pars{{1 \over t + 5} - {1 \over t - 5}}\,\dd t \\[3mm]&={1 \over 5}\,\ln\pars{\verts{t + 5 \over t - 5}} =\color{#66f}{\large% {1 \over 5}\,\ln\pars{\verts{\tan\pars{x/2} + 5 \over \tan\pars{x/2} - 5}}} + \mbox{a constant.} \end{align}

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