If the value of the integral int_{-1}^1 frac{ | vedclass

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(B) Let $I = \int_{-1}^{1} \frac{\cos \alpha x}{1+3^x} dx$  $(1)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we get:$I = \int

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$\frac{\pi}{2}$

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(B) Let $I = \int_{-1}^{1} \frac{\cos \alpha x}{1+3^x} dx$  $(1)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we get:$I = \int_{-1}^{1} \frac{\cos \alpha (-x)}{1+3^{-x}} dx = \int_{-1}^{1} \frac{\cos \alpha x}{1+\frac{1}{3^x}} dx = \int_{-1}^{1} \frac{3^x \cos \alpha x}{3^x+1} dx$  $(2)$Adding $(1)$ and $(2)$:$2I = \int_{-1}^{1} \frac{\cos \alpha x}{1+3^x} dx + \int_{-1}^{1} \frac{3^x \cos \alpha x}{1+3^x} dx$$2I = \int_{-1}^{1} \frac{(1+3^x) \cos \alpha x}{1+3^x} dx = \int_{-1}^{1} \cos \alpha x dx$Since $\cos \alpha x$ is an even function:$2I = 2 \int_{0}^{1} \cos \alpha x dx = 2 \left[ \frac{\sin \alpha x}{\alpha} \right]_{0}^{1} = \frac{2 \sin \alpha}{\alpha}$$I = \frac{\sin \alpha}{\alpha}$Given $I = \frac{2}{\pi}$,so $\frac{\sin \alpha}{\alpha} = \frac{2}{\pi}$.By inspection,$\alpha = \frac{\pi}{2}$ satisfies the equation since $\frac{\sin(\pi/2)}{\pi/2} = \frac{1}{\pi/2} = \frac{2}{\pi}$.

If the value of the integral $\int_{-1}^1 \frac{\cos \alpha x}{1+3^x} d x$ is $\frac{2}{\pi}$,then a value of $\alpha$ is

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