Animated Solution for Mathematics - Definite Integration: If the value of the integral ${\int }_{-1}^1\frac{\cos \alpha x}{1+3^x}dx$ is $\frac{2}{\pi }$. Then, a value of $\alpha$ is

Unlocking the Integral

My dear student, welcome to a beautiful journey through the world of definite integrals. Today, we are going to tackle a problem that looks intimidating at first glance, but reveals its elegance once we apply the right perspective.

We are looking at the integral:

We are told that this entire area equals $\frac{2}{\pi }$. Our mission is to uncover the mysterious value of $\alpha$.

Whenever you encounter an integral with symmetric limits, like our interval from $-1$ to $1$, your mathematical intuition should immediately scream for one specific tool: King's Property.

This property, ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$, is a lifesaver. In our case, $a=-1$ and $b=1$, so $a+b=0$. This means the property simplifies to replacing $x$ with $-x$.

We replace every $x$ in our integral with $-x$. The integral becomes:

We know that $\cos (\theta )$ is an even function, meaning $\cos (-\theta )=\cos (\theta )$. So, the numerator remains $\cos (\alpha x)$.

The denominator becomes $1+3^{-x}$. We can rewrite $3^{-x}$ as $\frac{1}{3^x}$. If we multiply the numerator and the denominator by $3^x$, the denominator transforms into $3^x+1$.

Suddenly, our transformed integral looks like this:

Now, let's add our original integral $I$ to our transformed integral $I$. We get:

Because the denominators are identical, we can combine the numerators:

If we factor out $\cos (\alpha x)$ from the numerator, we get $\cos (\alpha x)(1+3^x)$. This term perfectly cancels out the denominator, leaving us with the simple, elegant integral:

Since $\cos (\alpha x)$ is an even function, the area from $-1$ to $0$ is the same as the area from $0$ to $1$. Thus, $2I=2{\int }_0^1\cos (\alpha x)dx$, which simplifies to:

Integrating this is straightforward: the integral of $\cos (\alpha x)$ is $\frac{\sin (\alpha x)}{\alpha }$. Evaluating this from $0$ to $1$, we get:

We are left with the equation $\frac{\sin (\alpha )}{\alpha }=\frac{2}{\pi }$. By inspection, if we set $\alpha =\frac{\pi }{2}$, we get:

We have found our answer! The value is $\alpha =\frac{\pi }{2}$. The beauty of this problem lies in how the complexity of the exponential function was stripped away by the symmetry of the limits.