What is the integral of 1/(-csc(y)) ?

The integral of 1/(-csc(y)) is cos(y)+C

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integral of 1/(-csc(y))

\int \frac{1}{- csc ( y )} d y

cos (y) + C

Solution steps

\int \frac{1}{- csc ( y )} d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{csc ( y )} d y

Rewrite using trig identities

= - \int sin (y) d y

Use the common integral:

\int sin (y) d y = - cos (y)

= - (- cos (y))

Simplify

= cos (y)

Add a constant to the solution

= cos (y) + C

f (x) = - \frac{x + 3 arcsin ( x )}{4}

\frac{d}{d x} (e^{3 x} cos (3 x))

\int 19 sin^{2} (x co) s^{2} x d x

\int \frac{2 e ^{- \frac{1}{x}}}{x ^{2}} d x

x \to \infty lim (2 + 3 - 2)