Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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.

share|cite|improve this question
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 '14 at 22:05
Substituting $u=tan(x)$ and with some luck, you need to find a primitive of $$\frac{\sqrt{1+u^2}}{1+u^3}$$ – LeGrandDODOM Feb 20 '14 at 22:22
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?

share|cite|improve this answer
@yoda, how about this? – lab bhattacharjee Feb 21 '14 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 '14 at 14:16
@AméricoTavares, agreed & rectified. Thanks for your observation – lab bhattacharjee Feb 21 '14 at 15:11
@lab bhattacharjee. Thanks. – yoda Feb 21 '14 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.

share|cite|improve this answer

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

share|cite|improve this answer
@ Felix Martin.Thank you. – yoda Feb 21 '14 at 23:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.