What is the integral of 1/(sin^2(t)) ?

The integral of 1/(sin^2(t)) is -cot(t)+C

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integral of 1/(sin^2(t))

\int \frac{1}{sin ^{2} ( t )} d t

- cot (t) + C

Solution steps

\int \frac{1}{sin ^{2} ( t )} d t

Use the common integral:

\int \frac{1}{sin ^{2} ( t )} d t = - cot (t)

= - cot (t)

Add a constant to the solution

= - cot (t) + C

r = 1 0^{2 t}

\frac{\partial}{\partial x} (sin (π (4 x - 3 y)))

n = 1 \sum \infty \frac{( n )}{ln ( n )}

\frac{\partial}{\partial x} (x + y + ln (x y))

\int 1 - v d v

What is the integral of 1/(sin^2(t)) ?
The integral of 1/(sin^2(t)) is -cot(t)+C