If the value of the integral $\int\limits_{-1}^1 \frac{\cos \alpha x}{1+3^x} d x$ is $\frac{2}{\pi}$.Then, a value of $\alpha$ is
Solution
<p>$$\begin{aligned}
& \text { Given, } \int_\limits{-1}^1 \frac{\cos \alpha x}{1+3^x} d x=\frac{2}{\pi} \\
& \begin{aligned}
I & =\int_\limits{-1}^1 \frac{\cos \alpha x}{1+3^x} d x \\
\Rightarrow I & =\int_\limits0^1\left(\frac{\cos \alpha x}{1+3^x}+\frac{\cos \alpha x}{1+3^{-x}}\right) d x \\
& =\int_\limits0^1 \cos \alpha x d x \\
& =\left(\frac{\sin \alpha x}{\alpha}\right)_0^1 \\
& =\frac{\sin \alpha}{\alpha} \\
\Rightarrow & \frac{\sin \alpha}{\alpha}=\frac{2}{\pi} \\
\Rightarrow & \alpha=\frac{\pi}{2}
\end{aligned}
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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