So I have the following definite integral

$$\int_{\beta = 0}^{\beta = \pi}\int_{\alpha = 0}^{\alpha = 2\pi}\left(\frac{\sin\beta\cos\alpha\cos\beta}{\gamma + \sin^2\beta\cos^2\alpha}\right)^2 \mathrm d\alpha \,\mathrm d\beta $$

Personally I can not see an easy way of solving this, so I thought a series of substitutions might help. First I say that $x=\cos\alpha$, this yields

$$\int_{\beta = 0}^{\beta = \pi}\int_{x_1}^{x_2}\left(\frac{\sin\beta x\cos\beta}{\gamma + \sin^2\beta x^2}\right)^2 \frac{\mathrm dx}{-\sin\alpha} \, \mathrm d\beta $$

I then argue that because $1 - \cos^2 = sin^2$ I can say that $(1-x^2)^{1/2} = \sin\alpha$ yielding $$\int_{\beta = 0}^{\beta = \pi}\int_{x_1}^{x_2}\left(\frac{\sin\beta x\cos\beta}{\gamma + \sin^2\beta x^2}\right)^2 \frac{\mathrm dx}{-\sqrt{1-x^2}} \, \mathrm d\beta $$

I then use a similar argument, but starting with $y = \sin\beta$ to finally get

$$\int_{y_1}^{y_2}\int_{x_1}^{x_2}\left(\frac{y x(1-y^2)^{1/2}}{\gamma + y^2 x^2}\right)^2 \frac{\mathrm dx}{-\sqrt{1-x^2}} \frac{\mathrm dy}{\sqrt{1-y^2}} $$

Which, assuming my maths is correct, should come to

$$\int_{y_1}^{y_2}\int_{x_1}^{x_2} -\frac{y^2 x^2(1-y^2)^{1/2}}{(1-x^2)^{1/2}(\gamma + y^2 x^2)^2}\,\mathrm dx\, \mathrm dy $$

Now to me this looks more plausible to be solvable, however, I am at a lose as to what the new limits should be. From my understanding, I should just do $y_1 = \sin(\beta_1) \rightarrow y_1 = 0$, and $y_2 = \sin(\beta_2) \rightarrow y_2 = 0$, but those limits look like nonsense to me, as this would make the integral zero.

My question is, is there something I have done wrong here? Or an improper assumption? And can someone help find the new limits.


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