Question

integration -π to π sin(4x) cot(x/2) dx

Answer 1

2π

Answer 2

The integrand $f(x) = \sin(4x) \cot(x/2)$ is an even function because $f(-x) = \sin(-4x) \cot(-x/2) = (-\sin(4x))(-\cot(x/2)) = f(x)$. Therefore, the integral can be written as $2 \int_{0}^{\pi} \sin(4x) \cot(x/2) dx$. Using the identity $\cot(x/2) = \csc x + \cot x$, the integrand becomes $\sin(4x)(\csc x + \cot x)$. Expanding $\sin(4x) = 4\sin x \cos x \cos(2x)$ and simplifying, we get the integrand as a polynomial in $\cos x$: $8\cos^4 x + 8\cos^3 x - 4\cos^2 x - 4\cos x$. The integral is $2 \int_{0}^{\pi} (8\cos^4 x + 8\cos^3 x - 4\cos^2 x - 4\cos x) dx$. We use the following definite integrals: $\int_{0}^{\pi} \cos x dx = 0$ $\int_{0}^{\pi} \cos^2 x dx = \frac{\pi}{2}$ $\int_{0}^{\pi} \cos^3 x dx = 0$ $\int_{0}^{\pi} \cos^4 x dx = \frac{3\pi}{8}$ Substituting these values: $2 \left[ 8 \left(\frac{3\pi}{8}\right) + 8(0) - 4 \left(\frac{\pi}{2}\right) - 4(0) \right] = 2 [3\pi - 2\pi] = 2\pi$. The singularity at $x=0$ is removable as $\lim_{x\to 0} \sin(4x) \cot(x/2) = 8$.