# How to estimate the following integral: $\int_0^1 \frac{1-\cos x}{x}\,dx$

How to estimate the following integral? $$\int_0^1 \frac{1-\cos x}{x}\,dx$$

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Have you tried anything? – Brian Fitzpatrick Apr 22 '13 at 5:59
yes i did. i got sin^2x/ (x)(1+cosx) and then simplified into sinx/x and sinx/(1+cosx) and now im lost :( – parker Apr 22 '13 at 6:02

First note that the integral exists since $$0 \leq \dfrac{1-\cos(x)}x = \dfrac{2 \sin^2(x/2)}x \leq \dfrac{x}2$$ Hence, the integral is between $0$ and $1/4$. To compute the integral, proceed as follows. We have $$\cos(x) = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} \mp = \sum_{k=0}^{\infty}(-1)^k \dfrac{x^{2k}}{(2k)!}$$ Hence, $$1-\cos(x) = \dfrac{x^2}{2!} - \dfrac{x^4}{4!} + \dfrac{x^6}{6!} \pm = \sum_{k=1}^{\infty} (-1)^{k-1} \dfrac{x^{2k}}{(2k)!}$$ This gives us $$\dfrac{1-\cos(x)}x = \sum_{k=1}^{\infty} (-1)^{k-1} \dfrac{x^{2k-1}}{(2k)!}$$ Now lets get back to the integral. \begin{align} \int_0^1 \dfrac{1-\cos(x)}x dx & = \int_0^1 \sum_{k=1}^{\infty} (-1)^{k-1}\dfrac{x^{2k-1}}{(2k)!} dx = \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{(2k)!}\int_0^1 x^{2k-1} dx\\ & = \sum_{k=1}^{\infty}\dfrac{(-1)^{k-1}}{(2k)!} \cdot \dfrac1{2k} \end{align}

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can you show us step by steps??cause im confused :( – parker Apr 22 '13 at 6:10
Once you have this you can use the alternating series test to find a bound for the error in approximating by the $n$th partial sum. – Brian Fitzpatrick Apr 22 '13 at 6:11
you guys are so smart lol – parker Apr 22 '13 at 6:13
@parker Have added the intermediate details. – user17762 Apr 22 '13 at 6:15
@parker In the summation take the first $4$ terms, i.e., $$\dfrac1{2 \cdot 2!} - \dfrac1{4 \cdot 4!} + \dfrac1{6 \cdot 6!} - \dfrac1{8 \cdot 8!}$$ to get an accuracy of $1e-5$. – user17762 Apr 22 '13 at 6:31

Since $\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$, $1-\cos(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{2n}}{(2n)!}$, so $\frac{1-\cos(x)}{x} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{2n-1}}{(2n)!}$.

Therefore

\begin{align} \int_0^1 \frac{1-\cos(x)}{x} dx &= \int_0^1 dx \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{2n-1}}{(2n)!}\\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{(2n)!} \int_0^1 x^{2n-1} dx\\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{(2n)!} \frac1{2n} \\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{2n(2n)!}\\ \end{align}

This is an alternating series with terms decreasing in absolute value, so its sum is between any two consecutive terms.

The first two terms are $\frac1{4}$ and $\frac{-1}{4\cdot 4!} = \frac{-1}{96}$, so the result is slightly less than $\frac1{4}$.

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I Understood the computation of the integral. however when i try to find the estimate for the integral that is accurate to within 10^-5, i got 1/4-1/96+1/4320-1/322560+1/36288000 approximate to 0.2398117422. is 0.239811722 accurate to within 10^-5? im just confused again..

so my question would be

1st. did i computate the integral right?

2nd. is there other way to do the integral? not in terms of summation.

Correct me if im wrong please and tell me step by step

Thank you for your time and effort

Sincerely

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The error is always less than the last term used, so once you used 1/332560, the error is less than that term. – marty cohen Apr 23 '13 at 1:31