Si $\alpha$ es un ángulo agudo, demuestran que $\displaystyle \int_0^1 \frac{dx}{x^2+2x\cos{\alpha}+1} = \frac{\alpha}{2\sin{\alpha}}.$

Question

Si $\alpha$ es un ángulo agudo, muestran que $\displaystyle \int_0^1 \frac{dx}{x^2+2x\cos{\alpha}+1} = \frac{\alpha}{2\sin{\alpha}}.$

Mi intento:

Escribir $x^2+2x\cos{\alpha}+1 = (x+\cos{\alpha})^2+1-\cos^2{\alpha} = (x+\cos{\alpha})^2+\sin^2{\alpha}$, tenemos:

$\displaystyle \begin{aligned}\int_0^1 \frac{dx}{x^2+2x\cos{\alpha}+1} & = \int_0^1 \frac{dx}{(x+\cos{\alpha})^2+\sin^2{\alpha}} = \bigg[\frac{1}{\sin{\alpha}}\bronceado^{-1}\left(\frac{x+\cos{\alpha}}{\sin{\alpha}}\right)\bigg]_{x=0}^1 \\ & = \frac{1}{\sin{\alpha}}\bigg[\color{blue}{\tan^{-1}\left(\frac{1+\cos{\alpha}}{\sin{\alpha}}\right)-\tan^{-1}\left(\frac{\cos{\alpha}}{\sin{\alpha}}\right)}\bigg] \end{aligned}$

No estoy seguro, sin embargo, cómo el azul bits se reduce a $\frac{1}{2}\alpha$. Consejos/sugerencias? Gracias.

Answer 1

Use $1 + \cos \alpha = 2\cos ^ 2 (\frac{\alpha}{2})$

y $\sin \alpha = \cos \sin (\frac{\alpha}{2)} 2 (\frac{\alpha}{2})$

Answer 2

Dibujar el triángulo de la derecha para ver que $\arctan(\cot(\alpha)) = \frac {\pi} \alpha {2} $