I want help to solve this integral: L = ∫ cost √ [16sin²t + 1/(4sint)] dt, t = 0 to π?

I agree with Jeffrey -- another way to see it is to note that the substitution x = pi - t introduces a triple negative (it reverses (or negates) the order of integration, it negates the cosine factor, and it negates the differential).

HOWEVER -- you should always first check that your integral makes sense, or CONVERGES. Doubt is cast by the fact that the integral is IMPROPER: the integrand is discontinuous at both endpoints. We are lucky in this case: since u = sin(t) is small, 1/u dominates u^2, and so the integrand is similar to 1/sqrt(u), which converges on any finite interval. If, for example, the integrand were cos(t)*sqrt[16*sin^2(t) + 1/(16*sin^2(t))]*dt = sqrt[16u^2 + 1/16u^2]*du, then u^2 would again be dominated by 1/u^2, so the integrand would be similar to 1/sqrt[u^2] = 1/u, but this does NOT converge on any interval containing 0.

L= integral cost*sqrt(16sin^2t+ 1/(4sint)) dt

= cos(pi - t)*sqrt(16sin^2(pi-t) + 1/4sin(pi-t)) dt

= -cost*sqrt(16sin^2t + 1/4sint) dt

2L = 0

L = 0/2 = 0

int_0^pi cos(t)*sqrt{16*sin^2(t) + [1/(4*sin(t))]} dt

u = sin(t); du = cos(t) dt; a = 0; b = 0

int_0^0 sqrt[16(u^2) + (1/(4u))] du = 0