The value of ` int_(- pi//2) ^(pi//2) ( cos^2 x)/( 1+3^x ) dx ` is

Options

A. `pi/4`
B. `4 pi`
C. `(pi)/(2)`
D. `2pi`

Answer

To solve the integral

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2 x}{1 + 3^x} \, dx, \]

we can use the property of definite integrals that states:

\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx. \]

In our case, \( a = -\frac{\pi}{2} \) and \( b = \frac{\pi}{2} \). Thus, we have:

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2 x}{1 + 3^x} \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2\left(-x\right)}{1 + 3^{-x}} \, dx. \]

Since \(\cos(-x) = \cos x\), we can simplify this to:

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2 x}{1 + \frac{1}{3^x}} \, dx. \]

Next, we can rewrite the denominator:

\[ 1 + \frac{1}{3^x} = \frac{3^x + 1}{3^x}. \]

Thus, we have:

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2 x \cdot 3^x}{3^x + 1} \, dx. \]

Now we have two expressions for \(I\):

1. \( I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2 x}{1 + 3^x} \, dx \)
2. \( I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2 x \cdot 3^x}{3^x + 1} \, dx \)

Next, we can add these two expressions together:

\[ 2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{\cos^2 x}{1 + 3^x} + \frac{\cos^2 x \cdot 3^x}{3^x + 1} \right) \, dx. \]

Factoring out \(\cos^2 x\):

\[ 2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 x \left( \frac{1 + 3^x}{1 + 3^x} \right) \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 x \, dx. \]

Now we can compute the integral of \(\cos^2 x\):

Using the identity:

\[ \cos^2 x = \frac{1 + \cos 2x}{2}, \]

we can write:

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 x \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + \cos 2x}{2} \, dx = \frac{1}{2} \left( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos 2x \, dx \right). \]

Calculating the first integral:

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx = \pi. \]

The second integral evaluates to zero because \(\cos 2x\) is an even function over symmetric limits:

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos 2x \, dx = 0. \]

Thus, we have:

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 x \, dx = \frac{1}{2} \left( \pi + 0 \right) = \frac{\pi}{2}. \]

Now substituting back into our expression for \(2I\):

\[ 2I = \frac{\pi}{2} \implies I = \frac{\pi}{4}. \]

Therefore, the value of the integral is:

\[ \boxed{\frac{\pi}{4}}. \]