Integral of ln(sinx+cosx) with respect to x from -pi/4 to pi/4
To find the integral of ln(sin(x) + cos(x)) with respect to x from -π/4 to π/4, we can use the definite integral formula. However, the integral of ln(sin(x) + cos(x)) does not have a simple closed-form solution. Therefore, we will need to evaluate it numerically using approximation methods.
One approach is to use numerical integration techniques such as the trapezoidal rule or Simpson's rule. These methods approximate the area under the curve by dividing the interval of integration into smaller subintervals and approximating the area within each subinterval.
Here's how you can apply Simpson's rule to evaluate the integral:
1. Divide the interval [-π/4, π/4] into smaller subintervals. Let's say we choose n subintervals, where n is an even number for Simpson's rule. For simplicity, let's consider n = 4.
- Subinterval 1: [-π/4, -π/8]
- Subinterval 2: [-π/8, 0]
- Subinterval 3: [0, π/8]
- Subinterval 4: [π/8, π/4]
2. Approximate the integral within each subinterval using Simpson's rule formula:
∫[a,b] ln(sin(x) + cos(x)) dx ≈ (Δx/3) * [f(a) + 4f((a+b)/2) + f(b)]
where Δx = (b - a) / 2, f(x) = ln(sin(x) + cos(x)), a and b are the limits of each subinterval.
3. Calculate the approximate integral for each subinterval and sum them up:
Integral ≈ (Δx/3) * [f(a1) + 4f((a1+b1)/2) + f(b1)] +
(Δx/3) * [f(a2) + 4f((a2+b2)/2) + f(b2)] +
(Δx/3) * [f(a3) + 4f((a3+b3)/2) + f(b3)] +
(Δx/3) * [f(a4) + 4f((a4+b4)/2) + f(b4)]
where a1 = -π/4, b1 = -π/8, a2 = -π/8, b2 = 0, a3 = 0, b3 = π/8, a4 = π/8, b4 = π/4.
4. Simplify the expression and calculate the final numerical approximation of the integral.
Keep in mind that you can increase the number of subintervals (n) for better accuracy, although it will require more calculations. There are also more advanced numerical integration methods available, such as Gaussian quadrature, that can yield even more accurate results.
sin(x) + cos(x) = sqrt(2) sin(x + pi/4)
So, the integral can be written as:
pi/4 ln(2) + Integral of Log[sin(x)]dx from 0 to pi/2
Let's call the integral in here I:
I = Integral of Log[sin(x)]dx from 0 to pi/2
Then note that because sin is symmetric w.r.t. reflection about x = pi/2 integrating to pi will yield twice the value:
2 I = Integral of Log[sin(x)]dx from 0 to pi
Then substitute in this integral
x = 2 y:
2 I = 2 Integral of Log[sin(2y)]dy from 0 to pi/2 ----->
I = Integral of
(Log(2) + Log[cos(y)] +Log[sin(y)] )dy from 0 to pi/2 =
pi/2 Log(2) + 2 I
because the integral of log[cos(y)] is the same as the integral of
log[sin(y)], as cos(y) is sin(pi/2-y).
Solving for I gives:
I = -pi/2 log(2)
The original integral is thus
- pi/4 log(2)