You have not proved to me that the Integral of lnsinx w.r.t.x from 0 to pi/2 is equal to:Integral of lnsinx w.r.t.x from 0 to pi/4 plus Integral of lncosx w.r.t.x from 0 to pi/4.
graph sin x from 0 to pi/2
on the same graph plot cos x from 0 to pi/4
Note that cos x from pi/4 to pi/2 is identical to sin x from pi/4 to pi/2 with only a reflection.
Thus for every value of ln sin x from pi/4 to p/2 there is an equal value of ln cos x from 0 to pi/4. Therefore the areas under will be identical.
Damon: That is a good answer which now I really understand.I thank you very much.
To prove that the integral of ln(sinx) with respect to x from 0 to pi/2 is equal to the integral of ln(sinx) with respect to x from 0 to pi/4 plus the integral of ln(cosx) with respect to x from 0 to pi/4, we can use the property of symmetry in the integral.
Here's the step-by-step explanation:
Step 1: Write out the given equation:
∫[0 to π/2] ln(sinx) dx = ∫[0 to π/4] ln(sinx) dx + ∫[0 to π/4] ln(cosx) dx
Step 2: Use the property of symmetry:
Since ln(sinx) and ln(cosx) are symmetric functions about x = π/4, we can rewrite the integral on the right-hand side as a doubled integral over the interval [0 to π/4] by taking advantage of the symmetry property. Thus, we have:
∫[0 to π/2] ln(sinx) dx = 2∫[0 to π/4] ln(sinx) dx
Step 3: Simplify the equation:
By substituting the previous equation into the given equation, we get:
2∫[0 to π/4] ln(sinx) dx = ∫[0 to π/4] ln(sinx) dx + ∫[0 to π/4] ln(cosx) dx
Step 4: Combine like terms:
By subtracting ∫[0 to π/4] ln(sinx) dx from both sides, we can simplify it further:
∫[0 to π/4] ln(sinx) dx = ∫[0 to π/4] ln(cosx) dx
Step 5: Evaluate the integrals:
By evaluating the integrals on both sides, we can confirm that they are equal:
∫[0 to π/4] ln(sinx) dx = [π/4 * ln(sin(π/4))] - [0 * ln(sin(0))] = π/4 * ln(1/√2) - 0 = π/4 * (-ln(√2))
∫[0 to π/4] ln(cosx) dx = [π/4 * ln(cos(π/4))] - [0 * ln(cos(0))] = π/4 * ln(1/√2) - 0 = π/4 * (-ln(√2))
Both integrals evaluate to the same expression, π/4 * (-ln(√2)), which means they are equal.
Therefore, we have proven that the integral of ln(sinx) with respect to x from 0 to π/2 is equal to the integral of ln(sinx) with respect to x from 0 to π/4 plus the integral of ln(cosx) with respect to x from 0 to π/4.