Question

Let for \( f(x) = 7\tan^8 x + 7\tan^6 x - 3\tan^4 x - 3\tan^2 x \), \( I_1 = \int_0^{\frac{\pi}{4}} f(x)dx \) and \( I_2 = \int_0^{\frac{\pi}{4}} x f(x)dx \). Then \( 7I_1 + 12I_2 \) is equal to:

• A. \( 2\pi \)
• B. \( \pi \)
• C. 1
• D. 2

Hint:

For integrals involving trigonometric functions, a substitution such as \( t = \tan x \) can simplify the calculations significantly.

Answer 1

The Correct Option is C

The given function is: \[ f(x) = 7\tan^8 x + 7\tan^6 x - 3\tan^4 x - 3\tan^2 x \] We are given two integrals: \[ I_1 = \int_0^{\frac{\pi}{4}} f(x) \, dx \] \[ I_2 = \int_0^{\frac{\pi}{4}} x f(x) \, dx \] For \( I_1 \), let \( \tan x = t \). Then: \[ I_1 = \int_0^1 (7t^6 - 3t^2) dt = [t^7 - t^3]_0^1 = 1 - 1 = 0 \] \[ I_2 = \int_0^{\pi/4} x (7 \tan^6 x - 3 \tan^2 x) (\sec^2 x) dx \] \[ = [x (\tan^7 x - \tan^3 x)]_0^{\pi/4} - \int_0^{\pi/4} (\tan^7 x - \tan^3 x) dx \] \[ = 0 - \int_0^{\pi/4} \tan^3 x (\tan^4 x - 1) dx \] \[ = 0 - \int_0^{\pi/4} \tan^3 x (\tan^2 x - 1) (1 + \tan^2 x) dx \] Put $\tan x = t$ \[ = - \int_0^1 (t^5 - t^3) dt = - \left[ \frac{t^6}{6} - \frac{t^4}{4} \right]_0^1 = - \left( \frac{1}{6} - \frac{1}{4} \right) = - \left( \frac{2 - 3}{12} \right) = \frac{1}{12} \] \[ 7I_1 + 12I_2 = 7(0) + 12 \left( \frac{1}{12} \right) = 1 \] Thus: \[ 7I_1 + 12I_2 = 1 \]