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Evaluate: $$I=\int_{0}^{\frac{\pi}{2}} \ln(2468^{2} \cos^2x+990^2 \sin^2x) .dx$$ The suggested solution:

$$f(y) = \int_{0}^{\pi/2} ln( y^{2}cos^{2}x + sin^{2}x)$$

Here- $ \frac{2468}{990} = y$

$$f'(y) = 2y \int_{0}^{\pi/2}\frac{cos^{2}x}{sin^{2}x + y^{2}cos^{2}x}dx$$

$$= 2y \int_{0}^{\pi/2}\frac{dx}{tan^{2}x + y^{2}}$$

$$= 2y \int_{0}^{\pi/2}\frac{sec^{2}x - tan^{2}x }{tan^{2}x + y^{2}}dx$$

$$= 2y . \frac{1}{y} tan^{-1}( \frac{1}{y}) |_{0}^{\infty} -2y\frac{\pi}{2} + y^{2}f'(y)$$

$$f'(y) = \frac{\pi}{1 + y}$$

$$f(y) = \pi ln(1 + y) + c$$

$$ y = 1 , f(1) = 0 , c = -\pi ln2$$

$$ f(y) = \pi log(1 + y) -\pi ln2 $$

User1234
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  • The suggeste solution is one way forward. Of course, one could have placed the parameter $y$ in front of the sine function instead. But one need not use two parameters, one each in front of both the cosine and sine functions. – Mark Viola Jun 02 '15 at 19:39
  • You can use the property of the log as discussed by @cameronwilliams. Then, you should be able to proceed as suggested. – Mark Viola Jun 02 '15 at 20:01
  • OK. I provided a hint as requested. – Mark Viola Jun 02 '15 at 20:20

2 Answers2

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Why make things more complicated on yourself? If you have $y\cos^2 x + z\sin^2 x$, factor out $z$ to get $z\left(\frac{y}{z}\cos^2 x + \sin^2 x\right)$. After using properties of $\log$, you see that the factor of $z$ out front more or less doesn't matter and what does matter is $\frac{y}{z}$. Moreover, it doesn't matter what $y$ or $z$ are but their ratio so you might as well just consider the case your friend suggested.

Cameron Williams
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  • $\log(ab)=\log(a)+\log(b)$ – Paul Jun 02 '15 at 18:47
  • "Sir" is a little formal! I'm not that old yet haha. But on a serious note: $$ \log\left(z\left(\frac{y}{z}\cos^2 x + \sin^2 x\right)\right) = \log z + \log\left(\frac{y}{z}\cos^2 x + \sin^2 x\right).$$ The factor of $z$ out front gets uncoupled from the entire integral you're interested in because of the log property. – Cameron Williams Jun 02 '15 at 18:47
  • The reason is that the $z$ (by itself) doesn't couple with the term you're truly interested in. What is coupled to it is $\frac{y}{z}$. Also no worries about the sir! We're mathematicians. No need to be that formal here but it is appreciated. – Cameron Williams Jun 02 '15 at 18:55
  • You make a good point. I'm just not explaining myself well enough. My point is that the $z$ by itself isn't doing anything that you want it to do, so why keep it around? Why do you want to include another variable when you can quite nearly reduce to a simpler case that you're actually interested in anyway? – Cameron Williams Jun 02 '15 at 18:59
  • No. The integral you are interested in is the one with the trigonometric functions. After you break apart the two integrals, you have $\log z$ just floating around which isn't interacting directly with the other integral. Adding it is really only making things more difficult on yourself. K.I.S.S. (keep it super simple!). – Cameron Williams Jun 02 '15 at 19:03
  • You're making things needlessly complicated on yourself. I'm not sure that I can explain it any better than I have already. Sorry – Cameron Williams Jun 02 '15 at 19:17
  • I would have done it more or less the way your friend proposed and you wrote in your post. – Cameron Williams Jun 02 '15 at 19:23
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HINT:

Write

$$\log (2468 \cos^2x +990\sin^2x)=\log(990)+\log \left(\frac{2468}{990}\cos^2 x+\sin^2 x\right)$$

and evaluate the integral of $\log 990$ separately.

Then, proceed as in the suggested way forward and use $y=2468/990$ to evaluate at the end of the process.

Mark Viola
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