What is the integral of-sec(x)tan(x) ?

The integral of-sec(x)tan(x) is -sec(x)+C

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integral of-sec(x)tan(x)

\int - sec (x) tan (x) d x

- sec (x) + C

Solution steps

\int - sec (x) tan (x) d x

Take the constant out:

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

= - \int sec (x) tan (x) d x

Apply u-substitution

: \int 1 d u

= - \int 1 d u

Integral of a constant:

\int a d x = a x

= - 1 \cdot u

Substitute back

u = sec (x)

= - 1 \cdot sec (x)

Simplify

= - sec (x)

Add a constant to the solution

= - sec (x) + C

\frac{d}{d x} (ln (x + x^{2} - 9))

\frac{d}{d x} (\frac{x + 1}{x ^{3} + x - 2})

\frac{d}{d x} (x ln (5 x))

3 x^{2} - 4 x + 6, at x = 2

y^{^{''}} + 4 y = 0, y (0) = 1, y (1) = 0

What is the integral of-sec(x)tan(x) ?
The integral of-sec(x)tan(x) is -sec(x)+C