MHB Can the derivative of the given integral be simplified to -A?

[Suvadip]

Can it be proved?
$$\left(\frac{-2\sin A}{1-\cos A}\right)\cos\left(\frac{A}{2}\right)\tan^{-1}\left[\cos \left(\frac{A}{2}\right)\right]=\frac{\pi^2-4A^2}{8}$$

 

[Sudharaka]

Can it be proved?
$$\left(\frac{-2\sin A}{1-\cos A}\right)\cos\left(\frac{A}{2}\right)\tan^{-1}\left[\cos \left(\frac{A}{2}\right)\right]=\frac{\pi^2-4A^2}{8}$$

Hi suvadip, :)

This trigonometric identity is not valid. For example substituting \(A=\dfrac{\pi}{2}\) we get zero in the right hand side whereas \(-\dfrac{1}{\sqrt{2}}\tan^{-1}\dfrac{1}{\sqrt{2}}\neq 0\) on the right hand side.

 

[Suvadip]

Hi suvadip, :)

This trigonometric identity is not valid. For example substituting \(A=\dfrac{\pi}{2}\) we get zero in the right hand side whereas \(-\dfrac{1}{\sqrt{2}}\tan^{-1}\dfrac{1}{\sqrt{2}}\neq 0\) on the right hand side.

Please look at the attached sheet. How to reach the final answer

 

[Sudharaka]

Please look at the attached sheet. How to reach the final answer

Your attachment is not very clear and I cannot read it. If you can post a more clear attachment or write down the equations using LaTeX I would perhaps be able to help. :)

 

[Suvadip]

Your attachment is not very clear and I cannot read it. If you can post a more clear attachment or write down the equations using LaTeX I would perhaps be able to help. :)

I have to prove
$$\int_0^{\pi/2}\frac{log(1+cosA cosx)}{cosx}dx=\frac{\pi^2-4A^2}{8}$$

I have arrived at

$$\frac{d}{dA}(I)=-sinA \int_0^1\frac{2}{1+cosA +(1-cos A)z^2}dz$$ where I is the given integral. Am I correct so far and how to proceed at the answer from there? I have used the Leibnitz's rule for differentiation under the sign of integration.

 

[Saitama]

I have to prove
$$\int_0^{\pi/2}\frac{log(1+cosA cosx)}{cosx}dx=\frac{\pi^2-4A^2}{8}$$

I have arrived at

$$\frac{d}{dA}(I)=-sinA \int_0^1\frac{2}{1+cosA +(1-cos A)z^2}dz$$

How do you get this? After differentiating wrt A, I get,
$$\frac{dI}{dA}=\int_0^{\pi/2}\frac{-\sin A}{1+\cos A \cos x}dx$$
Can you integrate $\frac{dx}{1+a\cos x}$? Hint: Use $\cos x=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$.

EDIT: I see that you already tried that. Try this, it makes the algebra a bit easier to handle.
$$\frac{dI}{dA}=\int_0^{\pi/2}\frac{-\sin A}{1+\cos A \cos x}dx=\int_0^{\pi/2}\frac{-\sin A}{1+\cos A \sin x}dx$$
Use $\sin x=\frac{2\tan(x/2)}{1+\tan^2(x/2)}$ to get
$$\frac{dI}{dA}=-\sin A\int_0^{\pi/2} \frac{\sec^2(x/2)}{1+\tan^2(x/2)+2\cos A\tan(x/2)}dx$$
Use the substitution $\tan(x/2)=t$,
$$\Rightarrow \frac{dI}{dA}=-2\sin A\int_0^1 \frac{dt}{t^2+2t\cos A+1}=-2\sin A\int_0^1 \frac{dt}{(t+\cos A)^2+\sin^2A}$$
I suppose you can solve after this. :)

 

[Opalg]

I have to prove
$$\int_0^{\pi/2}\frac{\log(1+\cos A \cos x)}{\cos x}dx=\frac{\pi^2-4A^2}{8}$$

I have arrived at

$$\frac{d}{dA}(I)=-\sin A \int_0^1\frac{2}{1+\cos A +(1-\cos A)z^2}dz$$ where I is the given integral. Am I correct so far and how to proceed at the answer from there? I have used the Leibnitz's rule for differentiation under the sign of integration.

You are correct so far (having differentiated under the integral sign and then made the substitution $z = \tan(x/2)$). The next step is to write this as $$\frac{d}{dA}(I)=\frac{-2\sin A}{1-\cos A} \int_0^1\frac{1}{z^2 +\frac{1+\cos A}{1-\cos A}}dz.$$ Now use the fact that $$\frac{1+\cos A}{1-\cos A} = \cot^2(A/2)$$ to get $$\frac{d}{dA}(I)= \frac{-2\sin A}{1-\cos A} \int_0^1\frac{1}{z^2 +\cot^2(A/2)}dz = \frac{-2\sin A}{1-\cos A} \Bigl[\tan(A/2)\arctan\bigl(z\tan(A/2)\bigr)\Bigr]_0^1 = \frac{-2\sin A}{1-\cos A}\frac A2\tan(A/2)$$ (provided that $|A| < \pi$, so that $\arctan\bigl(\tan(A/2)\bigr) = A/2$). Using half-angle formulas again, you can write this as $$\frac{\frac{-2}{1+\tan^2(A/2)}}{1- \frac{1-\tan^2(A/2)}{1+\tan^2(A/2)}}A\tan(A/2) = \frac{-A\tan(A/2)}{\tan^2(A/2)} = -A\cot(A/2).$$ If that last expression was just $A$ (without the $\cot(A/2)$) then you could integrate it to get $-A^2/2$ which, together with the initial value $0$ when $A = \pi/2$, would give the formula that you are looking for.

But, unless I have made some silly mistake, that extra $\cot(A/2)$ is there, and that gives a function $-A\cot(A/2)$ that does not have an elementary integral.

 

[Saitama]

Hi Opalg! :)

I am not sure but I seem to be getting the final answer. If the integral I reached is evaluated further, I get
$$\frac{dI}{dA}=2\left(\tan^{-1}\left(\frac{\cos A}{\sin A}\right)-\tan^{-1}\left(\frac{1+\cos A}{\sin A}\right)\right)$$
Using the formula for $\tan^{-1}a-\tan^{-1}b$, I simplified it to
$$\frac{dI}{dA}=-A$$
Please check if I did anything wrong, thank you. :)

 

[Opalg]

Hi Opalg! :)

I am not sure but I seem to be getting the final answer. If the integral I reached is evaluated further, I get
$$\frac{dI}{dA}=2\left(\tan^{-1}\left(\frac{\cos A}{\sin A}\right)-\tan^{-1}\left(\frac{1+\cos A}{\sin A}\right)\right)$$
Using the formula for $\tan^{-1}a-\tan^{-1}b$, I simplified it to
$$\frac{dI}{dA}=-A$$
Please check if I did anything wrong, thank you. :)

I'm very willing to believe that you're right, but I'll leave it to the OP to sort out the details. (Wink)