Final answer:
The integral from 0 to π of dt/√t + sin(t) is tested for convergence by comparison to the well-known convergent integral of 1/√t.
The comparison test confirms convergence of the original integral.
Step-by-step explanation:
The problem asks to test for convergence of the integral from 0 to π of dt/√t + sin(t).
To approach this, we consider the behavior of the function as t approaches 0, since √t is undefined at t=0 and the integral may have a singularity there.
However, because the sine function is bounded, we will mainly focus on the √t term.
To test the convergence near t=0, we can compare to the integral of 1/√t, which is a well-known improper integral that converges (as it is of the form 1/t^p where p<1).
Since 1/(√t + sin(t)) ≥ 1/√t and 1/√t converges near zero, by the comparison test, our integral also converges near zero.
Throughout the rest of the interval [0, π], the integrand remains continuous and thus no other singularities exist that could affect convergence.