If the value of the integral ∫ cosαx / 1+3^x dx; x ∈ [-1, 1] is 2/π. Then, a value of α is

Question

If the value of the integral \(\int \limits _{-1}^1 \frac{\cos \alpha x}{1 + 3^x} dx\) is \(\frac{2}{\pi}\). Then, a value of \(\alpha\) is

(1) \(\frac{\pi}{6}\)

(2) \(\frac{\pi}{2}\)

(3) \(\frac{\pi}{3}\)

(4) \(\frac{\pi}{4}\)

Answer 1

Correct option is (2) \(\frac \pi 2\)

Let \(I=\int\limits_{-1}^{+1} \frac{\cos \alpha x}{1+3^{x}} d x\)

\(\mathrm{I}=\int\limits_{-1}^{+1} \frac{\cos \alpha \mathrm{x}}{1+3^{-x}} \mathrm{dx}\)

\(\left(\operatorname{using} \int\limits_{a}^{b} f(x) d x=\int\limits_{a}^{b} f(a+b-x) d x\right)\)

Add (I) and (II),

\(2 \mathrm{I}=\int\limits_{-1}^{+1} \cos (\alpha \mathrm{x}) \mathrm{dx}=2 \int\limits_{0}^{1} \cos (\alpha \mathrm{x}) \mathrm{dx}\)

\(\mathrm{I}=\frac{\sin \alpha}{\alpha}=\frac{2}{\pi} \quad\text{(given)}\)

\(\therefore \alpha=\frac{\pi}{2}\)