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$$ \mbox{Is it possible to calculate this integral}\quad \int{1 \over \cos^{3}\left(x\right) + \sin^{3}\left(x\right)}\,{\rm d}x\quad {\large ?} $$

I have tried $\dfrac{1}{\cos^3(x)+\sin^3(x)}$=$\dfrac{1}{(\cos(x)+\sin(x))(1-\cos x\sin x)}$ then I made a decomposition. But I'm still stuck. Thank you in advance.

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Alpha gets something that looks like it might help. There are some imaginary terms in there, so the initial result is not correct. –  Ross Millikan Feb 20 at 22:05
    
Substituting $u=tan(x)$ and with some luck, you need to find a primitive of $$\frac{\sqrt{1+u^2}}{1+u^3}$$ –  G.T.R Feb 20 at 22:22

3 Answers 3

up vote 2 down vote accepted

Where you have left of $$I=\int\frac1{(\cos x+\sin x)(1-\sin x\cos x)}=\int\frac{\cos x+\sin x}{(1+2\sin x\cos x)(1-\sin x\cos x)}$$

Let $\displaystyle\int(\cos x+\sin x)\ dx=\sin x-\cos x=u\implies u^2=1-2\sin x\cos x$

$$\implies I=\int\frac{2du}{(2-u^2)(1+u^2)}$$

Again, $\displaystyle\frac3{(2-u^2)(1+u^2)}=\frac{(2-u^2)+(1+u^2)}{(1+u^2)(2-u^2)}=\frac1{(1+u^2)}+\frac1{(2-u^2)}$

Finally use this for the second integral and the first one is too simple to be described, right?

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@yoda, how about this? –  lab bhattacharjee Feb 21 at 5:30
    
+1, There is a typo in the second integral: $$\frac1{(\cos x+\sin x)(1-\sin x\cos x)}=\frac{\cos x+\sin x}{(1+2\sin x\cos x)(1-\sin x\cos x)}.$$ Applying your substitution $u=\sin x-\cos x$ the integral $I$ becomes $$\int \frac{2du}{\left( 2-u^{2}\right) \left( u^{2}+1\right) }.$$ –  Américo Tavares Feb 21 at 14:16
1  
@AméricoTavares, agreed & rectified. Thanks for your observation –  lab bhattacharjee Feb 21 at 15:11
    
@lab bhattacharjee. Thanks. –  yoda Feb 21 at 21:11

The substitution $u = \tan(\frac{x}{2})$ converts any integrand that is a rational function in the two variables $\cos x$ and $\sin x$ into a rational function in $u,$ which can then be integrated by standard methods. See p. 56 of Hardy's The Integration of Functions of a Single Variable.

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

With $t \equiv x - \pi/4$: \begin{align} &{1 \over \bracks{\cos\pars{x} + \sin\pars{x}}\bracks{1 - \cos\pars{x}\sin\pars{x}}} ={\root{2} \over \cos\pars{t}\bracks{2 - \cos\pars{2t}}} ={\root{2} \over \cos\pars{t}\braces{2 - \bracks{2\cos^2\pars{t} - 1}}} \\[3mm]&={\root{2} \over \cos\pars{t}\bracks{3 - 2\cos^2\pars{t}}} ={\root{2} \over 2}\, {1 \over \cos\pars{t}\bracks{\root{3}/2 - \cos\pars{t}}\bracks{\root{3}/2 + \cos\pars{t}}} \\[3mm]&={\root{2} \over 2}\bracks{% {4/3\over \cos\pars{t}} + {3/2 \over \root{3}/2 - \cos\pars{t}} + {3/2 \over \root{3}/2 + \cos\pars{t}}} \\[3mm]&={2\root{2} \over 3}\,{1 \over \cos\pars{t}} +{3\root{2} \over 4}\bracks{% {1 \over \root{3}/2 - \cos\pars{t}} + {1 \over \root{3}/2 + \cos\pars{t}} } \end{align}

$$ \int{\dd t \over \cos\pars{t}}=\ln\pars{\sec\pars{t} + \tan\pars{t}} +\quad \mbox{a constant} $$

The remaining integrals can be easily performed with $s \equiv \tan\pars{t/2}$.

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@ Felix Martin.Thank you. –  yoda Feb 21 at 23:35

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