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I've tried to evaluate the following integral:

$$ \int \frac{\sin{x}}{\cos{x}\sqrt{\cos^2{x}+\cos{x}+1}}\mathrm dx $$

But I don't know what to do after the substitution $u=\cos{x}$:

$$-\int \frac{1}{u\sqrt{u^2+u+1}}\mathrm du$$

Please help.

Integreek
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tizzzato
  • 91

5 Answers5

3

Substitute $u+\frac12=\frac{\sqrt3}2\tan t$ \begin{aligned} &\int{\frac{1}{u\sqrt{u^{2}+u+1}}}du\\ =& -\int{\frac{1}{\cos(t+\frac\pi3)}}dt = -\int\frac{d\left[\sin(t+\frac\pi3) \right]}{1-\sin^2(t+\frac\pi3) }\\ =&\ \frac12\ln\frac{1-\sin(t+\frac\pi3)}{1+\sin(t+\frac\pi3)} = \frac12\ln \frac{\sqrt{u^{2}+u+1}-\frac12u-1}{\sqrt{u^{2}+u+1}+\frac12u+1} \\ \end{aligned}

Quanto
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\iverson}[1]{\left[\left[\,{#1}\,\right]\right]} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\int{\sin\pars{x} \over \cos\pars{x}\root{\cos^{2}\pars{x} + \cos\pars{x} + 1}} \dd x} \\[5mm] & \sr{u\ =\ \cos\pars{x}}{=} \ -\int{\dd u \over u\root{u^{2} + u + 1}} \dd x \\[5mm] & \sr{t\ =\ \root{\ds{u^{2} + u + 1}} - u}{=} \quad 2\int{\dd t \over 1 - t^{2}} = \ln\pars{1 + t \over 1 - t} \\[5mm] = & \ \ln\pars{1 + \root{\ds{u^{2} + u + 1}} - u \over 1 - \root{\ds{u^{2} + u + 1}} + u} \\[5mm] = & \ \bbx{\color{#44f}{\begin{array}{l} \ds{\ln\pars{1 + \root{\cos^{2}\pars{x} + \cos\pars{x} + 1} - \cos\pars{x} \over 1 - \root{\cos^{2}\pars{x} + \cos\pars{x} + 1} + \cos\pars{x}}} \\ +\ \mbox{( a constant )} \end{array}}} \end{align}

Felix Marin
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\begin{aligned}\int{\frac{\mathrm{d}u}{u\sqrt{u^{2}+u+1}}}&=\int{\frac{\mathrm{d}u}{u\sqrt{\left(u+\frac{1}{2}\right)^{2}+\frac{3}{4}}}}\\ &=\int{\frac{\frac{2}{\sqrt{3}}\,\mathrm{d}u}{u\sqrt{1+\left(\frac{2u+1}{\sqrt{3}}\right)^{2}}}}\\ &=2\int{\frac{\mathrm{d}y}{\left(\sqrt{3}y-1\right)\sqrt{1+y^{2}}}}\\ &=2\int{\frac{\mathrm{d}\varphi}{\sqrt{3}\sinh{\varphi}-1}}\end{aligned}

Can you continue from here ?

CHAMSI
  • 9,222
  • Unfortunately, we haven't studied hyperbolic functions, so we are supposed to solve this integral without $sinh$. – tizzzato Jan 07 '23 at 16:12
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    Okay. Then, you could do as @Quanto suggested, use $ y = \tan{\varphi} $ instead of $ y=\sinh{\varphi} $. – CHAMSI Jan 07 '23 at 18:42
0

Continuing from your $u$-integral, substitute

$$t = \frac{\sqrt{u^2+u+1}-1}u \implies u = \frac{2t-1}{1-t^2} \implies du = \frac{2(t^2-t+1)}{(1-t^2)^2} \, dt$$

so that

$$- \int \frac{du}{u\sqrt{u^2+u+1}} = - \int \frac{1-t^2}{(2t-1)\left(\frac{2t^2-t}{1-t^2}+1\right)} \, \frac{2(t^2-t+1)}{(1-t^2)^2} \,dt = -2 \int \frac{dt}{2t-1}$$

user170231
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0

The fastest way to evaluate the integral you were stuck at is to take $|u|$ common from the square root and perform the substitution $t=\frac1u$:

$$\begin{align}\int\frac{\mathrm du}{u\sqrt{u^2+u+1}}&=\text{sgn}(u)\int\frac{\mathrm du}{u^2\sqrt{\frac1{u^2}+\frac1u+1}}\\&\overset{t=\frac1u}{=}\text{sgn}(u)\int\frac{\mathrm dt}{\sqrt{t^2+t+1}}\\&=\text{sgn}(u)\sinh^{-1}\left(\frac{u+2}{\sqrt3u}\right)+C\end{align}$$

Integreek
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