For each of the following improper integrals, note Why it is omproper, and then set up the limit(s) that would be used to evaluate the integral.

${\int}_{1}^{\mathrm{\infty}}\frac{1}{{x}^{3}}dx$

Jason Farmer
2021-09-29
Answered

For each of the following improper integrals, note Why it is omproper, and then set up the limit(s) that would be used to evaluate the integral.

${\int}_{1}^{\mathrm{\infty}}\frac{1}{{x}^{3}}dx$

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lobeflepnoumni

Answered 2021-09-30
Author has **99** answers

For the given integration the upper limit is infinity also the integrand of the limit goes to infinity in the range of given integration at point x = 0.

Therefore, the given integral is improper.

Now to set up the given integral:

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By considering,

$\int {\mathrm{sin}}^{n}\left(x\right){\mathrm{cos}}^{3}\left(x\right)dx$

Prove that:

$\frac{{\mathrm{sin}}^{8}\left(x\right)}{8}-\frac{{\mathrm{sin}}^{6}\left(x\right)}{6}+\frac{1}{24}=\frac{{\mathrm{cos}}^{8}\left(x\right)}{8}-\frac{{\mathrm{cos}}^{6}\left(x\right)}{3}+\frac{{\mathrm{cos}}^{4}\left(x\right)}{4}$

I have integrated the suggested integral successfully, that being:

$\frac{{\mathrm{sin}}^{n+1}\left(x\right)}{n+1}-\frac{{\mathrm{sin}}^{n+3}\left(x\right)}{n+3}+c$

I see some resemblance but I fail to connect and finish the proof. Any hints on finding the relation between the powers of sin and cos through this integral? Obviously we could expand${\mathrm{sin}}^{2}\left(x\right)=(1-{\mathrm{cos}}^{2}\left(x\right))$ but this seems far too tedious.

Prove that:

I have integrated the suggested integral successfully, that being:

I see some resemblance but I fail to connect and finish the proof. Any hints on finding the relation between the powers of sin and cos through this integral? Obviously we could expand

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