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

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

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integral of cos(x)sec(x)tan(x)

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

- ln ∣ cos (x) ∣ + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int - \frac{1}{u} d u

= \int - \frac{1}{u} d u

Take the constant out:

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

= - \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = cos (x)

= - ln ∣ cos (x) ∣

Add a constant to the solution

= - ln ∣ cos (x) ∣ + C

\int_{0}^{2} 2 x - 1 d x

9 x y^{^{'}} - 2 y = - \frac{x ^{2}}{y ^{8}}

\frac{\partial}{\partial x} (lo g_{y} (x))

\frac{\partial}{\partial x} (x^{3} y + 7 x^{2} y^{2})

f (x) = \frac{x}{( x ^{2} - 1 ) ^{4}}

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