Question

5-16 Find the general indefinite integral.

Answer 1

Simplify the integrand

$\dfrac{1-\sin^2(t)}{\sin^2(t)} = \dfrac{\cos^2(t)}{\sin^2(t)} = \cot^2(t)$

Recall the derivative of csc(t),

$\dfrac d{dt} \csc(t) = -\cot^2(t)$

It follows that

$\displaystyle \int \frac{1-\sin^2(t)}{\sin^2(t)} \, dt = \int \cot^2(t) \, dt = \boxed{-\csc(t) + C}$