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The following integration problem appears in our calculus assignment:

$$ \int \limits_{-\pi /2014}^{\pi /2014}\dfrac{1}{2014^{x}+1}\left(\dfrac{\sin ^{2014}x}{\sin ^{2014}x+\cos ^{2014}x}\right) dx .$$

But the problem is I have no idea how to begin this problem. Could anyone give me some help ?

Any hints/ideas are much appreciated.

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  • $\begingroup$ This is possibly an extension of the integration rules surrounding $$ \int_a^b\frac{dx}{1+\cot^{n}\left(x\right)} $$ since the second fraction within the integrand can be re-written to get $$ \frac{1}{1+\cot^{n}\left(x\right)}? $$ $\endgroup$ – jm324354 Feb 7 '15 at 2:51
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    $\begingroup$ Make the $u$-substitution $u=-x$ and see what happens. $\endgroup$ – Sameer Kailasa Feb 7 '15 at 2:52
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    $\begingroup$ Less than $$5.942350125 \cdot 10^{-5660}.$$ $\endgroup$ – user64494 Feb 7 '15 at 17:37
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let $$f(x)=\dfrac{\sin^{2014}{x}}{\sin^{2014}{x}+\cos^{2014}{x}}\Longrightarrow f(x)=f(-x)$$ $$I=\int_{-a}^{a}\dfrac{f(x)}{1+2014^x}dx=\int_{-a}^{a}\dfrac{f(-x)}{1+2014^{-x}}dx=\int_{-a}^{a}\dfrac{f(x)}{1+2014^{-x}}$$ so $$2I=\int_{-a}^{a}f(x)\left(\dfrac{1}{1+2014^x}+\dfrac{1}{1+2014^{-x}}\right)dx=\int_{-a}^{a}f(x)dx$$ where $a=\dfrac{\pi}{2014}$ so we only find $$I'=\int_{-\frac{\pi}{2014}}^{\frac{\pi}{2014}}\dfrac{\sin^{2014}{x}}{\sin^{2014}{x}+\cos^{2014}{x}}dx$$ if $\dfrac{\pi}{2014}$ replace $\dfrac{\pi}{2}$,we have simple result because $$I''=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{\sin^{2014}{x}}{\sin^{2014}{x}+\cos^{2014}{x}}dx=2\int_{0}^{\frac{\pi}{2}}\dfrac{\sin^{2014}{x}}{\sin^{2014}{x}+\cos^{2014}{x}}dx=2I'''$$ and let $x\to \frac{\pi}{2}-x$,then we have $$I'''=\int_{0}^{\frac{\pi}{2}}\dfrac{\cos^{2014}{x}}{\sin^{2014}{x}+\cos^{2014}{x}}dx$$ so $$2I'''=\int_{0}^{\frac{\pi}{2}}\dfrac{\sin^{2014}{x}+\cos^{2014}{x}}{\sin^{2014}{x}+\cos^{2014}{x}}dx=\int_{0}^{\frac{\pi}{2}}1dx=\frac{\pi}{2}$$

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    $\begingroup$ +1. Outstanding answer and not something most students would have thought of! The odd-even nature of sine and cos is not taught irrelevantly-this kind of computation demonstrates that most clearly. The littlest details can be critical in a problem. $\endgroup$ – Mathemagician1234 Feb 7 '15 at 17:53
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Let $\displaystyle I = \int \limits_{-\pi /2014}^{\pi /2014}\dfrac{1}{2014^{x}+1}\left( \dfrac{\sin ^{2014}x}{\sin ^{2014}x+\cos ^{2014}x}\right) dx .$

Applying $\displaystyle \int_a^bf(x)\;{dx} = \int_a^b f(a+b-x)\;{dx}$, so we have

$\displaystyle I = \int \limits_{-\pi /2014}^{\pi /2014}\dfrac{1}{2014^{-x}+1}\left( \dfrac{\sin ^{2014}x}{\sin ^{2014}x+\cos ^{2014}x}\right) dx $

Adding and noting that $\frac{1}{2014^{x}+1}+\frac{1}{2014^{-x}+1} = 1$ we have

$\begin{aligned} \displaystyle 2I & = \int \limits_{-\pi /2014}^{\pi /2014} \left(\dfrac{1}{2014^{-x}+1}+\frac{1}{2014^{-x}+1}\right)\left( \dfrac{\sin ^{2014}x}{\sin ^{2014}x+\cos ^{2014}x}\right) dx \\& = \int \limits_{-\pi /2014}^{\pi /2014} \left( \dfrac{\sin ^{2014}x}{\sin ^{2014}x+\cos ^{2014}x}\right) dx = 2\int \limits_{0}^{\pi /2014} \left( \dfrac{\sin ^{2014}x}{\sin ^{2014}x+\cos ^{2014}x}\right)\;{dx}\end{aligned} $

because an integral of an even function over $[a, -a]$ is twice over $[0, a]$; thus

$\begin{aligned} \displaystyle I & = \int \limits_{0}^{\pi /2014} \left( \dfrac{\sin ^{2014}x}{\sin ^{2014}x+\cos ^{2014}x}\right)\;{dx}\end{aligned} $

Whoever created the question was trying to create something slightly different, I bet.

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  • $\begingroup$ You have a typo. In "Adding and noting that...." you have 2004 (twice) instead of 2014. Not that it makes any difference to the argument. $\endgroup$ – DanielWainfleet Nov 28 '18 at 10:55
  • $\begingroup$ @DanielWainfleet Thanks. $\endgroup$ – Caddyshack Jan 6 at 1:30

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