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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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The correct answer is:
$\frac{\pi}{2}$
Let $I=\int_{-1}^{+1} \frac{\cos \alpha x}{1+3^x} d x$ ...(I)
$I=\int_{-1}^{+1} \frac{\cos \alpha x}{1+3^{-x}} d x$
$\left(u \operatorname{sing} \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right)$ ...(II)
Add (1) and (II)
$\begin{aligned}
& 2 I=\int_{-1}^{+1} \cos (\alpha x) d x=2 \int_0^1 \cos (\alpha x) d x \\
& I=\frac{\sin \alpha}{\alpha}=\frac{2}{\pi}(\text { given }) \\
& \therefore \alpha=\frac{\pi}{2}
\end{aligned}$
$I=\int_{-1}^{+1} \frac{\cos \alpha x}{1+3^{-x}} d x$
$\left(u \operatorname{sing} \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right)$ ...(II)
Add (1) and (II)
$\begin{aligned}
& 2 I=\int_{-1}^{+1} \cos (\alpha x) d x=2 \int_0^1 \cos (\alpha x) d x \\
& I=\frac{\sin \alpha}{\alpha}=\frac{2}{\pi}(\text { given }) \\
& \therefore \alpha=\frac{\pi}{2}
\end{aligned}$
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