What is the integral of 1/(sec^2(y)) ?

The integral of 1/(sec^2(y)) is 1/2 (y+1/2 sin(2y))+C

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integral of 1/(sec^2(y))

\int \frac{1}{sec ^{2} ( y )} d y

\frac{1}{2} (y + \frac{1}{2} sin (2 y)) + C

Solution steps

\int \frac{1}{sec ^{2} ( y )} d y

Rewrite using trig identities

= \int \frac{1 + cos ( 2 y )}{2} d y

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 + cos (2 y) d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int 1 d y + \int cos (2 y) d y)

\int 1 d y = y

\int cos (2 y) d y = \frac{1}{2} sin (2 y)

= \frac{1}{2} (y + \frac{1}{2} sin (2 y))

Add a constant to the solution

= \frac{1}{2} (y + \frac{1}{2} sin (2 y)) + C

\frac{d}{d x} ((4 - x^{\frac{2}{3}})^{3})

x \to - 8 lim (1 + - 2 x)

\frac{\partial}{\partial x} (x - 2)

\int y sin (x + y^{2}) d y

y^{^{'}} = 3 y t

What is the integral of 1/(sec^2(y)) ?
The integral of 1/(sec^2(y)) is 1/2 (y+1/2 sin(2y))+C