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

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


  • 3
    $\begingroup$ 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. $\endgroup$ – Fredrik Meyer Dec 19 '11 at 6:06
  • $\begingroup$ 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. $\endgroup$ – Kyris Dec 19 '11 at 6:20
  • $\begingroup$ I don't have any solution with the integrals, need some hints... $\endgroup$ – Kyris Dec 19 '11 at 6:22
  • $\begingroup$ 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 $x^3$. $\endgroup$ – André Nicolas Dec 19 '11 at 6:26
  • $\begingroup$ Try wolphram alpha :D wolframalpha.com/input/?i=%28x^-5%29*%28sin+x%29 Paste the whole link, it will solve all your problems :) $\endgroup$ – nb1 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.

  • $\begingroup$ Do we have anything else with the (2)? I tried but can't find any idea... $\endgroup$ – Kyris Dec 20 '11 at 5:08

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.