3

1) $\displaystyle \int_{-\pi}^\pi \frac{\sin^3(x)dx}{x^2}$

2) $\displaystyle \int_{0}^{\pi/2} \frac{\cos^3(x)\sin(x)dx}{1 + \cos^2x}$

I have no ideas how to deal with it. I tried to evaluate them by parts and use change of variables, but nothing works. Thanks in advance!

Alex
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Michael
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3 Answers3

3

For the 1st one , note that you're integrating an odd function over $[-\pi,\pi]$.

As David Mitra pointed out, the function is bounded($\lim_{x\to 0}\frac{\sin^3(x)}{x^2}=0$) , and continuous everywhere except 0 (and hence improperly integrable), allowing me to use the oddness of the function .

Ishan Banerjee
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  • It should also be pointed out that the integrand is bounded (thus integrable). – David Mitra Feb 20 '13 at 17:02
  • Ah, yes. I forgot to mention that. – Ishan Banerjee Feb 20 '13 at 17:04
  • I do not think that noting that it is an odd function is quite enough, since we have a singularity at $x=0$. For instance you cannot argue in that manner for $\int_{-1}^1\frac{1}{x}=0$. You can argue that the Cauchy principle value is zero however. But David Mitra deals with this issue by realizing that the singularity is removable. – Baby Dragon Feb 20 '13 at 17:13
2

There is a pretty way to solve the second integral. After substituting $t= \cos x$ denote

$$ I=\int_{0}^{1} \frac{t^3 dt}{1+t^2}\\ J=\int_{0}^{1}\frac{t dt}{1+t^2} = -\frac{\log 2}{2}\\ I+J=\int_{0}^{1}\frac{(t^3+t)dt}{1+t^2}=\int_{0}^{1}t dt= - \frac{1}{2} $$ Hence $I=\frac{\log 2-1}{2}$

Alex
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1

The second one may be done by substituting $u = \cos{x}$, $du = -\sin{x} dx$. You then get

$$\int_0^1 du \frac{u^3}{1+u^2}$$

You may substitute again by letting $v=u^2$, $dv = 2 u \, du$ and get

$$\int_0^1 dv \frac{v}{1+v} = \frac{1}{2}(1-\log{2})$$

Ron Gordon
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