Question

Let \( I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\tan^2 x}{1+5^x} \, dx \). Then:

• A. \( I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^2 x \, dx \)
• B. \( 2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^2 x \, dx \)
• C. \( I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{1+5^x} \, dx \)
• D. \( 2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 5 \tan^2 x \, dx \)
• E. \( 2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{1+5^x} \, dx \)

Hint:

Utilize symmetry and properties of even and odd functions to simplify integrals, especially when working with trigonometric identities and exponential functions.

Answer 1

The Correct Option is B

Start by recognizing the symmetry properties of the integrand: \[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\tan^2 x}{1+5^x} \, dx \] Notice that \( \tan^2(-x) = \tan^2 x \), which implies that \( \tan^2 x \) is an even function. 
However, \( 5^x \) is not symmetric around \( x = 0 \). Let's examine the function under a substitution that utilizes this symmetry: \[ u = -x, \quad dx = -du \] \[ \int_{\frac{\pi}{4}}^{-\frac{\pi}{4}} \frac{\tan^2 u}{1+5^{-u}} \, (-du) \] \[ = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\tan^2 x}{1+\frac{1}{5^x}} \, dx \] \[ = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{5^x \tan^2 x}{5^x+1} \, dx \] Now, using the symmetry of \( 5^x \) and \( \frac{1}{5^x} \): \[ I + I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^2 x \, dx \] \[ 2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^2 x \, dx \] This shows that the integral of the original function, multiplied by two, equals the integral of \( \tan^2 x \) over the same interval, confirming that the correct answer is: \[ 2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^2 x \, dx \]