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Complete the integrals

  1. $\displaystyle\int \frac{\sin(x)dx}{\sin(x)+\cos(x)}$

  2. $\displaystyle\int_{0}^{\infty } x^{-5}\sin(x)dx$

Find the following limit

$\displaystyle\lim_{x \to \infty } \frac{x^{3}+\sin^{3}x}{x^{3}+\cos^{3}x}$

Thanks!

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    Two things: 1) Your title is not very helpful. 2) Try to describe what you have tried yourself, where you're stuck and so on.2011-12-19
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    In the limit above, i try to divide x^3 and get trouble with sin^3(x)/x^3 and cos^3(x)/x^3 from x to inf.2011-12-19
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    I don't have any solution with the integrals, need some hints...2011-12-19
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    You almost did the limit. Divide top and bottom by $x^3$. Note that $\frac{\sin^3 x}{x^3}$ and $\frac{\cos^3 x}{x^3}$ both approach $0$, since our trig functions stay between $-1$ and $1$, get crushed on division by $x^3$.2011-12-19
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    Try wolphram alpha :D http://www.wolframalpha.com/input/?i=%28x^-5%29*%28sin+x%29 Paste the whole link, it will solve all your problems :)2011-12-19
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    I tried but not understand it... For example with the 1st integral, i used and it make me crazy :(2011-12-19
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    I dont think so..just type "integral ((sin x)/((sin x)+(cos x)))" and after the answer pops up, cluck on the tab "show steps".2011-12-19
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    it's very long... i think there will be another way to make it more simple.2011-12-19
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    @Kiris: It can be made shorter, but only with a trick. Are you sure it is not a *definite* integral that is asked for?2011-12-19

1 Answers 1

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HINTS:

For (1): Multiply the integrand by $\dfrac{\sin x-\cos x}{\sin x-\cos x}$ and use the double and half angle formulas to get a fraction containing only sines and cosines of $2x$ instead of $x$. If you do it right, the denominator will be a single trig function of $2x$, so you’ll be able to divide it through to get an integrand with no fractions.


For (2): I don’t see a straightforward integration, but the following indirect argument will work. Let $f(x)=x^{-5}\sin x$. Then $f(x)<0$ only when $(2n+1)\piwhy? It follows that the parts of the function below the $x$-axis contribute only a finite amount to the integral.

On the other hand, there is a positive $a<\pi/2$ such that if $0why?) and hence $\dfrac{\sin x}{x^5}\ge\dfrac1{2x^4}$. Thus, if $0

$$\int_b^a\frac{\sin x}{x^5}dx\ge\frac12\int_b^ax^{-4}dx=-\frac16\left[x^{-3}\right]_b^a=\frac16\left[x^{-3}\right]_a^b=\frac16\left(\frac1{b^3}-\frac1{a^3}\right)\;.$$ What is the limit of this as $b\to 0^+$? What can you conclude about $\displaystyle\int_0^\infty f(x)dx$?


For (3): This is very straightforward. How would you handle $$\lim_{x\to\infty}\frac{x^3+1}{x^3-2}\;?$$ There’s a standard technique, and it works equally well on this problem. Remember, $-1\le \sin x,\cos x\le 1$, and you have the squeeze theorem to work with.

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    Do we have anything else with the (2)? I tried but can't find any idea...2011-12-20