Multiple choice

Let $\displaystyle \alpha ,\beta $ be the distincet positive roots of the equation $\displaystyle \tan x=2x$ then evaluate $\displaystyle \int_{0}^{1}\left ( \sin \alpha x.\sin \beta x \right )dx$ independent of $\displaystyle \alpha ,\beta $.

  1. 0

  2. -2

  3. 1

  4. 2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For each root, tan x = 2x implies sin x = 2x cos x. Thus sin(alpha - beta)/(alpha - beta) and sin(alpha + beta)/(alpha + beta) are both equal to 2 cos(alpha)cos(beta). Substitution into the product-to-sum integral makes the two terms cancel, giving 0.

AI explanation

The definite integral of sin(alpha x) multiplied by sin(beta x) from 0 to 1 is given by the formula (sin(alpha - beta))/(2(alpha - beta)) minus (sin(alpha + beta))/(2(alpha + beta)). Because alpha and beta are roots of tan x = 2x, we can replace 2 with tan(alpha)/alpha and tan(beta)/beta. This gives the sine addition formulas as sin(alpha + beta) = sin(alpha) + sin(beta) and sin(alpha - beta) = sin(alpha) - sin(beta). Substituting these into the integral formula and cancelling terms yields 0. The result is 0.

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