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lfdahl
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Prove, that the definite integral
$$\int_{0}^{\pi}\ln (2-2\cos x)dx = 0.$$
$$\int_{0}^{\pi}\ln (2-2\cos x)dx = 0.$$
June29 said:Call the integral $I$. Let $x \mapsto \pi-x$ then we get $\displaystyle \int_0^{\pi}\ln(2+2\cos{x})\,{dx}$.
Then $\displaystyle 2I = \int_0^{\pi}\log(4\sin^2{x})\,{dx} \implies I = \int_0^{\pi}\log(\sin{x}) +\log(2)\,{dx}$
But $\displaystyle\log (\sin x)=-\sum\limits_{j=1}^{\infty }{\frac{\cos (2jx)}{j}}-\log( 2)$. Using this we end up with
$\displaystyle I = -\int_0^{\pi} \sum_{j=1}^{\infty} \frac{1}{j} \cos{2jx} \,{dx}= -\sum_{j=1}^{\infty} \frac{1}{j} \int_0^{\pi} \cos(2jx) \,{dx} = 0$ as required.
The purpose of this challenge is to find the value of x that makes the expression ∫ln(2−2cosx)dx equal to 0. This involves solving for the antiderivative of the given function and then plugging in the limits of integration to evaluate the definite integral.
The antiderivative of ln(2−2cosx) can be found by using integration techniques such as substitution or integration by parts. It is important to note that ln(2−2cosx) is an indefinite integral, so the resulting antiderivative will have a constant of integration (C) attached to it.
The limits of integration for this challenge are not specified, so it is up to the solver to choose appropriate limits based on the given function and the desired value of 0. Generally, the limits should be chosen such that the resulting integral is solvable.
Yes, there are two main considerations when solving this challenge: 1) the natural logarithm function becomes undefined for x = 0, and 2) the given function is periodic with a period of 2π. These factors should be taken into account when choosing the limits of integration and solving for the antiderivative.
The solution to this challenge represents the value of x that makes the given function's definite integral equal to 0. This value can have practical applications in various fields such as physics, engineering, and economics. Additionally, solving this challenge can help improve problem-solving skills and deepen understanding of integration techniques.