∫cot(u) du = ln|sin(u)| + C ∫sec(u) du = ln|sec(u) + tan(u)| + C ∫cosec(u) du = ln|cosec(u) + cot(u)| + C ∫sec2(u) du = tan(u) + C ∫cosec2(u) du = -cot(u) + C ∫sec(u) tan(u) du =sec(u) + C ∫cosec(u) cot(u) du = - cosec(u...
In summary, the given integration problem can be simplified by making the substitution \sin y = u, leading to the partial fractions \frac{A}{u-2} + \frac{B}{u+3}. This makes it easier to solve for the coefficients A and B. ...
I am actually looking for the expectation of x for the wavefunction that is Ψ(x)=2LSin(πxL) for 0<x<L. To do this I need to find the solution to this integral: f=∫0LΨ∗xΨdx=∫0Lx∗(2LSin(πxL))2dx The Attempt at a Solution Using Mathematica, I know that the answer...
Integral of csc xis, ∫ cosec x dx = ln |cosec x - cot x| + C (or) - ln |cosec x + cot x| + C (or) ln | tan (x/2) | + C Integral of sec xis, ∫ sec x dx = ln |sec x + tan x| + C (or) (1/2) ln | (1 + sin x) / (1 - sin x) (or) ln | ...