Integral of theta^3*lnsintheta with respect to theta from 0 to pi
∫[0,pi] x^3 ln(sin(x)) dx
you got me. wolframalpha . com figures the value, but it's not evaluated using elementary functions.
Integral [0,pi/2] of sin(x)ln(sinx) w.r.to x.
To find the integral of the function theta^3 * ln(sin(theta)) with respect to theta from 0 to pi, you can use the technique of integration by parts. Here's how you can proceed:
First, let's write the given integral:
∫[0 to pi] theta^3 * ln(sin(theta)) d(theta)
To use integration by parts, we need to choose two parts of the integrand, u and dv, such that the derivative of u is easy to integrate and the integral of dv is easy to compute. In our case, we can choose:
u = ln(sin(theta)) (and du = (cos(theta)/sin(theta)) d(theta))
dv = theta^3 d(theta) (and v = (1/4)theta^4)
Now, let's apply integration by parts formula:
∫ u dv = uv - ∫ v du
Plugging in the values of u, dv, du, and v, we get:
∫[0 to pi] ln(sin(theta)) * theta^3 d(theta)
= (1/4)theta^4ln(sin(theta)) ∣ [0 to pi] - ∫[0 to pi] (1/4)theta^4 * (cos(theta)/sin(theta)) d(theta)
Evaluating the definite integral:
= (1/4)pi^4ln(sin(pi)) - (1/4)0^4ln(sin(0)) - (1/4) ∫[0 to pi] (theta^4 * cos(theta))/sin(theta) d(theta)
Now, let's simplify the expression:
Since sin(pi) = sin(0) = 0, the first two terms in the expression cancel out. Hence the integral reduces to:
∫[0 to pi] (theta^4 * cos(theta))/sin(theta) d(theta)
This integral does not have a simple closed form solution and requires more advanced techniques like integration by parts or substitution using trigonometric identities. Hence, we cannot find the exact value without using numerical methods or approximation techniques.