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

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

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integral of 1/(-sec^2(x))

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

- \frac{1}{2} (x + \frac{1}{2} sin (2 x)) + C

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

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

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 x) d x

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 x + \int cos (2 x) d x)

\int 1 d x = x

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

= - \frac{1}{2} (x + \frac{1}{2} sin (2 x))

Add a constant to the solution

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

(4 x^{3} + 4 x y - 1) d x = (1 - 2 x^{2} - 2 y) d y

\frac{d}{d x} (∣ x ∣^{∣ x ∣})

x \frac{d y}{d x} + y = - 9 x y^{2}

\frac{\partial}{\partial x} (\frac{8}{x ^{2}})

\int \frac{1}{1 - 6 x} d x

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