The value of the definite integral int_0^1 fr | vedclass

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(D) We have the integral $I = \int_0^1 \frac{dx}{x^2 + 2x\cos \alpha + 1}$.Completing the square in the denominator: $x^2 + 2x\cos \alpha + \cos^2 \al

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$\frac{\alpha}{2 \sin \alpha}$

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(D) We have the integral $I = \int_0^1 \frac{dx}{x^2 + 2x\cos \alpha + 1}$.Completing the square in the denominator: $x^2 + 2x\cos \alpha + \cos^2 \alpha + 1 - \cos^2 \alpha = (x + \cos \alpha)^2 + \sin^2 \alpha$.Thus,$I = \int_0^1 \frac{dx}{(x + \cos \alpha)^2 + \sin^2 \alpha}$.Using the standard integral $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$,we get:$I = \left[ \frac{1}{\sin \alpha} 	an^{-1}\left( \frac{x + \cos \alpha}{\sin \alpha} ight) ight]_0^1$.Evaluating at the limits:$I = \frac{1}{\sin \alpha} \left[ 	an^{-1}\left( \frac{1 + \cos \alpha}{\sin \alpha} ight) - 	an^{-1}\left( \frac{\cos \alpha}{\sin \alpha} ight) ight]$.Using trigonometric identities $\frac{1 + \cos \alpha}{\sin \alpha} = \cot(\frac{\alpha}{2})$ and $\frac{\cos \alpha}{\sin \alpha} = \cot \alpha$:$I = \frac{1}{\sin \alpha} \left[ 	an^{-1}(\cot \frac{\alpha}{2}) - 	an^{-1}(\cot \alpha) ight]$.Since $	an^{-1}(\cot 	heta) = \frac{\pi}{2} - 	heta$:$I = \frac{1}{\sin \alpha} \left[ (\frac{\pi}{2} - \frac{\alpha}{2}) - (\frac{\pi}{2} - \alpha) ight] = \frac{1}{\sin \alpha} [\frac{\alpha}{2}] = \frac{\alpha}{2 \sin \alpha}$.

The value of the definite integral $\int_0^1 \frac{dx}{x^2 + 2x\cos \alpha + 1}$ for $0 < \alpha < \pi$ is equal to

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