Mathematics

The value of the integral $$\displaystyle\int^1_0\dfrac{dx}{x^2+2x\cos \alpha +1}$$, where $$0 < \alpha < \dfrac{\pi}{2}$$, is equal to


ANSWER

$$\dfrac{\alpha}{2\sin \alpha}$$


SOLUTION
$$\displaystyle\int^1_0\dfrac{dx}{x^2+2x\cos \alpha +1}$$

$$\implies \displaystyle\int^1_0\dfrac{dx}{x^2+2x\cos \alpha +\cos^2 \alpha+\sin^2 \alpha}$$

$$\implies \displaystyle \int_0^1 \cfrac{dx}{(x+ \cos \alpha)^2 + \sin^2 \alpha}$$

$$\implies \displaystyle \cfrac{1}{\sin \alpha} \Bigg[\tan^{-1} \cfrac{x+\cos x}{\sin x}\Bigg]_0^1 $$   $$\left[\because \int \cfrac{1}{x^2+a^2}=\dfrac{1}{a}tan^{-1}(\dfrac{x}{a}) \right]$$

$$\implies \displaystyle \cfrac{1}{\sin \alpha} \Bigg[\tan^{-1} \cfrac{1+\cos \alpha}{\sin \alpha} - \tan^{-1} \cfrac{\cos \alpha}{\sin \alpha}\Bigg] $$

$$\implies \cfrac{1}{\sin \alpha} \Bigg[ \tan^{-1} \cfrac{\cfrac{1+\cos \alpha}{\sin \alpha} - \cfrac{\cos \alpha}{\sin \alpha}}{1+\cfrac{\cos \alpha + \cos^2 \alpha}{\sin^2 \alpha}}\Bigg]$$

$$\implies \cfrac{1}{\sin \alpha} \tan^{-1} \cfrac{\sin \alpha}{1+\cos \alpha}$$

$$\implies \cfrac{1}{\sin \alpha} \tan^{-1} \tan \cfrac{\alpha}{2}$$

$$\implies \cfrac{\alpha}{2 \sin \alpha}$$
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