Is it the integral of lnsinx w.r.t x from 0 to pi/2 = integral lnsinx w.r.t x from 0 to pi/4 + integral lncosx w.r.t x from 0 to pi/4 ?
To determine whether the given equality is true or false, we need to evaluate both sides of the equation and compare the results.
Let's start by evaluating the integral of ln(sin(x)) with respect to x from 0 to π/2:
∫[0,π/2] ln(sin(x)) dx
To solve this integral, we can use integration by parts. The formula for integration by parts is:
∫ u dv = uv - ∫ v du
Let's assign u = ln(sin(x)) and dv = dx. Then, we can find du and v:
du = (1/sin(x)) * cos(x) dx
v = x
Now, applying the integration by parts formula, we have:
∫[0,π/2] ln(sin(x)) dx = uv - ∫[0,π/2] v du
∫[0,π/2] ln(sin(x)) dx = x * ln(sin(x)) - ∫[0,π/2] x * (1/sin(x)) * cos(x) dx
To solve the remaining integral, we need to use the substitution u = sin(x), du = cos(x) dx:
∫[0,π/2] ln(sin(x)) dx = x * ln(sin(x)) - ∫[0,π/2] x * (1/u) du
∫[0,π/2] ln(sin(x)) dx = x * ln(sin(x)) - ∫[0,1] x/u du
Evaluating the definite integral, we get:
∫[0,π/2] ln(sin(x)) dx = (π/2) * ln(sin(π/2)) - ∫[0,1] x/u du
∫[0,π/2] ln(sin(x)) dx = (π/2) * ln(1) - ∫[0,1] x/u du
∫[0,π/2] ln(sin(x)) dx = (π/2) * 0 - ∫[0,1] x/u du
∫[0,π/2] ln(sin(x)) dx = 0
On the right-hand side of the equation, let's evaluate the individual integrals:
∫[0,π/4] ln(sin(x)) dx:
Using the same steps as above, we can evaluate this integral:
∫[0,π/4] ln(sin(x)) dx = (π/4) * ln(sin(π/4)) - ∫[0,1] x/u du
∫[0,π/4] ln(sin(x)) dx = (π/4) * ln(1/√2) - ∫[0,1] x/u du
∫[0,π/4] ln(sin(x)) dx = (π/4) * ln(1/√2) - ∫[0,1] x/u du
Similarly, let's evaluate the other integral on the right-hand side:
∫[0,π/4] ln(cos(x)) dx:
Using integration by parts again, we assign u = ln(cos(x)) and dv = dx:
du = -(1/cos(x)) * sin(x) dx
v = x
∫[0,π/4] ln(cos(x)) dx = x * ln(cos(x)) - ∫[0,π/4] x * (−(1/cos(x))) * sin(x) dx
∫[0,π/4] ln(cos(x)) dx = x * ln(cos(x)) + ∫[0,π/4] x * sin(x)/cos(x) dx
Using the substitution u = cos(x), we get:
∫[0,π/4] ln(cos(x)) dx = x * ln(cos(x)) + ∫[1,√2/2] ln(u) du
Evaluating the definite integral:
∫[0,π/4] ln(cos(x)) dx = (π/4) * ln(cos(π/4)) + ∫[1,√2/2] ln(u) du
∫[0,π/4] ln(cos(x)) dx = (π/4) * ln(√2/2) + ∫[1,√2/2] ln(u) du
Now, let's sum up the two integrals:
∫[0,π/4] ln(sin(x)) dx + ∫[0,π/4] ln(cos(x)) dx = [(π/4) * ln(1/√2) - ∫[0,1] x/u du] + [(π/4) * ln(√2/2) + ∫[1,√2/2] ln(u) du]
∫[0,π/4] ln(sin(x)) dx + ∫[0,π/4] ln(cos(x)) dx = (π/4) * (ln(1/√2) + ln(√2/2)) - ∫[0,1] x/u du + ∫[1,√2/2] ln(u) du
Note that ln(1/√2) + ln(√2/2) = 0, so the equation becomes:
∫[0,π/4] ln(sin(x)) dx + ∫[0,π/4] ln(cos(x)) dx = 0 - ∫[0,1] x/u du + ∫[1,√2/2] ln(u) du
∫[0,π/4] ln(sin(x)) dx + ∫[0,π/4] ln(cos(x)) dx = - ∫[0,1] x/u du + ∫[1,√2/2] ln(u) du
Therefore, we can conclude that:
∫[0,π/2] ln(sin(x)) dx ≠ ∫[0,π/4] ln(sin(x)) dx + ∫[0,π/4] ln(cos(x)) dx