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$$\int_{0}^{1}\left(2x\sin \frac{1}{x}-\cos \frac{1}{x}\right)dx$$

Am stuck in this question. Can't solve by applying any of the properties of definite integral. What should one do? By performing indefinite integration and then putting limits but that would be quite lengthy I guess.

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Hint: Use the substitution $x\mapsto\frac1x$: $$ \begin{align} \int_0^1\left(2x\sin\left(\frac1x\right)-\cos\left(\frac1x\right)\right)\mathrm{d}x &=\int_1^\infty\frac{2\sin(x)-x\cos(x)}{x^3}\,\mathrm{d}x \end{align} $$ then notice $$ \frac{\mathrm{d}}{\mathrm{d}x}\frac{\sin(x)}{x^2}=\frac{x\cos(x)-2\sin(x)}{x^3} $$

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Hint: set $$ f(x) = x^2\sin \frac1x $$ and find $f'(x)$.

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  • 2
    $\begingroup$ For anyone knowing this function, that pops up as example in some instances, this is clear. Yet for someone that doesn't know it the question is "how to reach this if I don't know the primitive beforehand?" $\endgroup$ – DonAntonio Mar 7 '16 at 8:54
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    $\begingroup$ @Joanpemo The only thing I can say is that the integrand sure looks like the result of applying the product rule, so I tried a few different things that seemed like they might work. The third one I tried was this one. $\endgroup$ – Arthur Mar 7 '16 at 8:55
  • $\begingroup$ @Joanpemo I was halfway through that comment before you posted yours, so I really understand the concern. Many times it seems like you just have to know a lot of antiderivatives to do integration. While that may be true in a lot of other cases, in this case it did narrow down the search considerably to look at the general form of the expression. $\endgroup$ – Arthur Mar 7 '16 at 9:09
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    $\begingroup$ Also, one gets a bit genre savvy once one has seen enough of these problems. In this case, they have one problem with two terms rather than two problems with each of the terms separately. That says that the two terms should interact. One has a sine of some argument, the other has a cosine. That speaks in favour of the product rule, where one has been differentiated and the other hasn't. The difference between the terms is a polynomial factor, and those are the nicest to integrate and differentiate. That leaves trial and error. $\endgroup$ – Arthur Mar 7 '16 at 9:10
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By parts,

$$\int 2x\sin\left(\frac1x\right)dx=x^2\sin\left(\frac1x\right)dx-\int -\frac{x^2}{x^2}\cos\left(\frac1x\right)dx=x^2\sin\left(\frac1x\right)+\int \cos\left(\frac1x\right)dx.$$

You can conclude from there.

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  • $\begingroup$ Can we compute $\int\cos\left(\frac1x\right)\,\mathrm{d}x$ without using special functions? $\endgroup$ – robjohn Mar 7 '16 at 9:13
  • $\begingroup$ @robjohn: Wolfram will tell you. But here you don't need that. $\endgroup$ – Yves Daoust Mar 7 '16 at 9:16
  • $\begingroup$ Does it not use $\operatorname{Si}(x)$? $\endgroup$ – robjohn Mar 7 '16 at 9:18
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    $\begingroup$ @robjohn: here you don't need that. Read the question. $\endgroup$ – Yves Daoust Mar 7 '16 at 9:19

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