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

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

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

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

- cot (x) + C

Solution steps

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

Use the common integral:

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

= - cot (x)

Add a constant to the solution

= - cot (x) + C

x \to 2 + lim (\frac{x + 1}{x - 2})

\int_{0}^{π} 5 sin^{2} (t co) s^{4} t d t

f (x) = (ln (x))^{4}, at x = 2

y = \frac{x ^{3}}{5 - x ^{2}}

\frac{\partial}{\partial x} (ln (x^{2} + y^{4}))

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