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Titan97

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## Homework Statement

Prove that $$\int_{-\pi/2}^{\pi/2}\frac{log(1+b\sin x)}{\sin x}dx=\pi \arcsin(b)$$

Where ##|b|\le1##

## Homework Equations

$$\frac{d}{dx}\big(\int_a^bf(t)dt\big)=\int_a^b\frac{\partial f(t)}{\partial x}dx$$

## The Attempt at a Solution

Let $$f(b)=\int_{-\pi/2}^{\pi/2}\frac{log(1+b\sin x)}{\sin x}dx$$

Using Leibnitz theorem, $$f'(b)=\int_{-\pi/2}^{\pi/2}\frac{1}{1+b\sin x}dx$$

Now, ##\sin(x)=\frac{2\tan(x/2)}{1+\tan^2(x/2)}##

After simplifying and substituting ##\tan(x/2)=u##, I got

$$f'(b)=\frac{2}{\sqrt{1-b^2}}\big[\arctan{\frac{\sqrt{1+b}}{\sqrt{1-b}}}+\arctan{\frac{\sqrt{1-b}}{\sqrt{1+b}}}\big]$$

Now, for ##b=0##, its clear that ##f(b)=0##.

So $$f(b)=\frac{2}{\sqrt{1-b^2}}\big[\arctan{\frac{\sqrt{1+b}}{\sqrt{1-b}}}+\arctan{\frac{\sqrt{1-b}}{\sqrt{1+b}}}\big]$$

After putting ##b=\cos\phi## since ##|b|\le 1##, I got ##f(b)=\frac{\pi}{\sin({\arccos(b)})}## which is wrong.

Also, the new ##f(b)## does not satisfy ##f(0)=0##

Is my method correct?

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