What is the integral of 1/(4sec(y)) ?

The integral of 1/(4sec(y)) is 1/4 sin(y)+C

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integral of 1/(4sec(y))

\int \frac{1}{4 sec ( y )} d y

\frac{1}{4} sin (y) + C

Solution steps

\int \frac{1}{4 sec ( y )} d y

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{sec ( y )} d y

Rewrite using trig identities

= \frac{1}{4} \cdot \int cos (y) d y

Use the common integral:

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

= \frac{1}{4} sin (y)

Add a constant to the solution

= \frac{1}{4} sin (y) + C

\frac{d y}{d x} = \frac{x x ^{2} - 1}{y}

\int (3 p^{2} - 2)^{2} e^{- 4 p} d p

2 x^{2} + 2 x + 3

\frac{d}{d x} ((1 + x) e^{x})

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

What is the integral of 1/(4sec(y)) ?
The integral of 1/(4sec(y)) is 1/4 sin(y)+C