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yanalaym [24]
3 years ago
7

5-16 Find the general indefinite integral.

Mathematics
1 answer:
NeX [460]3 years ago
5 0

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}

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