11
$\begingroup$

Solve the following indefinite integrals:

$$ \begin{align} &(1)\;\;\int\frac{1}{\sin^3 x+\cos^3 x}dx\\ &(2)\;\;\int\frac{1}{\sin^5 x+\cos^5 x}dx \end{align} $$

My Attempt for $(1)$:

$$ \begin{align} I &= \int\frac{1}{\sin^3 x+\cos ^3 x}\;dx\\ &= \int\frac{1}{\left(\sin x+\cos x\right)\left(\sin^2 x+\cos ^2 x-\sin x \cos x\right)}\;dx\\ &= \int\frac{1}{\left(\sin x+\cos x\right)\left(1-\sin x\cos x\right)}\;dx\\ &= \frac{1}{3}\int \left(\frac{2}{\left(\sin x+\cos x\right)}+\frac{\left(\sin x+\cos x \right)}{\left(1-\sin x\cos x\right)}\right)\;dx\\ &= \frac{2}{3}\int\frac{1}{\sin x+\cos x}\;dx + \frac{1}{3}\int\frac{\sin x+\cos x}{1-\sin x\cos x}\;dx \end{align} $$

Using the identities

$$ \sin x = \frac{2\tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}},\;\cos x = \frac{1-\tan ^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}} $$

we can transform the integral to

$$I = \frac{1}{3}\int\frac{\left(\tan \frac{x}{2}\right)^{'}}{1-\tan^2 \frac{x}{2}+2\tan \frac{x}{2}}\;dx+\frac{2}{3}\int\frac{\left(\sin x- \cos x\right)^{'}}{1+(\sin x-\cos x)^2}\;dx $$

The integral is easy to calculate from here.

My Attempt for $(2)$:

$$ \begin{align} J &= \int\frac{1}{\sin^5 x+\cos ^5 x}\;dx\\ &= \int\frac{1}{\left(\sin x+\cos x\right)\left(\sin^4 x -\sin^3 x\cos x+\sin^2 x\cos^2 x-\sin x\cos^3 x+\cos^4 x\right)}\;dx\\ &= \int\frac{1}{(\sin x+\cos x)(1-2\sin^2 x\cos^2 x-\sin x\cos x+\sin^2 x\cos^2 x)}\;dx\\ &= \int\frac{1}{\left(\sin x+\cos x\right)\left(1-\sin x\cos x-\left(\sin x\cos x\right)^2\right)}\;dx \end{align} $$

How can I solve $(2)$ from this point?

$\endgroup$
1
  • $\begingroup$ Have you tried Maple or Mathematica? $\endgroup$ – Mhenni Benghorbal Dec 6 '13 at 7:09
3
$\begingroup$

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\large\tt\mbox{Just a hint:}$ Write $$ \int{\cos\pars{x}\,\dd x \over \cos\pars{x}\sin^{3}\pars{x} + \cos^{4}\pars{x}} = \int{\dd z \over \root{1 - z^{2}}z^{3} + \bracks{1 - z^{2}}^{2}} \quad\mbox{with}\quad z \equiv \sin\pars{x} $$ Use an Euler substitution: $\root{1 - z^{2}} \equiv t + \ic z$ which yields $1 - z^{2} = t^{2} + 2t\ic z - z^{2}$ such that $\ds{z = {1 - t^{2} \over 2t\ic}}$: \begin{align} \root{1 - z^{2}}&=t + {1 - t^{2} \over 2t} = {1 + t^{2} \over 2t} \\[3mm] \dd z&= {\pars{-2t}\pars{2t\ic} - \pars{2\ic}\pars{1 - t^{2}} \over -4t^{2}}\,\dd t = \ic\,{t^{2} + 1 \over 2t^{2}}\,\dd t \end{align} \begin{align} \int&=\int{1 \over \bracks{\pars{1 + t^{2}}/2t}\bracks{\pars{1 - t^{2}}/2t}^{3}\pars{-1/\ic} + \bracks{\pars{1 + t^{2}}/2t}^{4}} \,\ic\,{t^{2} + 1 \over 2t^{2}}\,\dd t \\[3mm]&=-8\int{t^{2} \over -\pars{1 - t^{2}}^{3} + \ic\pars{1 + t^{2}}^{3}}\,\dd t \end{align}

$\endgroup$
5
$\begingroup$

Given $$\displaystyle \int\frac{1}{\sin^5 x+\cos^5 x}dx$$

First we will simplify $$\sin^5 x+\cos^5 x = \left(\sin^2 x+\cos^2 x\right)\cdot \left(\sin^3 x+\cos^3 x\right) - \sin ^2x\cdot \cos ^2x\left(\sin x+\cos x\right)$$

$$\displaystyle \sin^5 x+\cos^5 x= (\sin x+\cos x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)$$

So Integral is $$\displaystyle \int\frac{1}{\sin^5 x+\cos^5 x}dx $$

$$\displaystyle = \int\frac{1}{(\sin x+\cos x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$$

$$\displaystyle = \int \frac{(\sin x+\cos x)}{(\sin x+\cos x)^2\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$$

$$\displaystyle = \int \frac{(\sin x+\cos x)}{(1+\sin 2x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$$

Let $$(\sin x-\cos x) = t\;,$$ Then $$(\cos +\sin x)dx = dt$$ and $$(1-\sin 2x) = t^2\Rightarrow (1+\sin 2x) = (2-t^2)$$

So Integral Convert into $$\displaystyle = 4\int\frac{1}{(2-t^2)\cdot(5-t^4)}dt = 4\int\frac{1}{(t^2-2)\cdot (t^2-\sqrt{5})\cdot (t^2+\sqrt{5})}dt$$

Now Using partial fraction, we get

$$\displaystyle = 4\int \left[\frac{1}{2-t^2}+\frac{1}{(2-\sqrt{5})\cdot 2\sqrt{5}\cdot (\sqrt{5}-t^2)}+\frac{1}{(2+\sqrt{5})\cdot 2\sqrt{5}\cdot (\sqrt{5}+t^2)}\right]dt$$

$$ = \displaystyle \sqrt{2}\ln \left|\frac{\sqrt{2}+t}{\sqrt{2}-t}\right|+\frac{1}{(2-\sqrt{5})\cdot 5^{\frac{3}{4}}}\cdot \ln \left|\frac{5^{\frac{1}{4}}+t}{5^{\frac{1}{4}}-t}\right|+\frac{2}{(2+\sqrt{5})\cdot 5^{\frac{3}{4}}}\cdot \tan^{-1}\left(\frac{t}{5^{\frac{1}{4}}}\right)+\mathbb{C}$$

where $$t=(\sin x-\cos x)$$

$\endgroup$
1
$\begingroup$

I am not sure how you can continue either (the second term in the denominator can be expressed as $1-\sin(2 x)/2 - \sin^2(2x)/4,$ but I am not aware of any double angle formula for $\sin x + \cos x.$ The simplest approach to your integral is to use the feared $u = \tan \frac{x}2$ substitution, which reduces the integral to a rational function integral....

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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