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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

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


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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. – Fredrik Meyer Dec 19 '11 at 6:06
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. – Kyris Dec 19 '11 at 6:20
I don't have any solution with the integrals, need some hints... – Kyris Dec 19 '11 at 6:22
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 $x^3$. – André Nicolas Dec 19 '11 at 6:26
Try wolphram alpha :D^-5%29*%28sin+x%29 Paste the whole link, it will solve all your problems :) – NikBels Dec 19 '11 at 6:30


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 $(2n+1)\pi<x<(2n+2)\pi$ 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 $0<x\le a$, $\dfrac{\sin x}x\ge\dfrac12$ (why?) and hence $\dfrac{\sin x}{x^5}\ge\dfrac1{2x^4}$. Thus, if $0<b<a$ 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... – Kyris Dec 20 '11 at 5:08

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