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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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    @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)\pi for some integer $n$. On the interval from $(2n+1)\pi$ to $(2n+2)\pi$, $|f(x)|\le(2n+1)^{-5}\pi^{-5}$, so $\int_{(2n+1)\pi}^{(2n+2)\pi}|f(x)|dx\le \frac1{(2n+1)^5\pi^5}\int_{(2n+1)\pi}^{(2n+2)\pi}dx=\frac1{(2n+1)^5\pi^4}\;,$ and $\sum_{n=0}^\infty\int_{(2n+1)\pi}^{(2n+2)\pi}|f(x)|dx\le\frac1{\pi^4}\sum_{n=0}^\infty\frac1{(2n+1)^5}\;.$ This sum is finite; why? 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 $0, $\dfrac{\sin x}x\ge\dfrac12$ (why?) and hence $\dfrac{\sin x}{x^5}\ge\dfrac1{2x^4}$. Thus, if $0 we have

$\int_b^a\frac{\sin x}{x^5}dx\ge\frac12\int_b^ax^{-4}dx=-\frac16\left[x^{-3}\right]_b^a=\frac16\left[x^{-3}\right]_a^b=\frac16\left(\frac1{b^3}-\frac1{a^3}\right)\;.$ 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