What is the integral of (sin(x)+tan(x))/(cos(x)) ?

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

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integral of (sin(x)+tan(x))/(cos(x))

\int \frac{sin ( x ) + tan ( x )}{cos ( x )} d x

- (ln ∣ cos (x) ∣ - sec (x)) + C

Solution steps

\int \frac{sin ( x ) + tan ( x )}{cos ( x )} d x

Apply u-substitution

= \int - \frac{u + 1}{u ^{2}} d u

Take the constant out:

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

= - \int \frac{u + 1}{u ^{2}} d u

Expand

\frac{u + 1}{u ^{2}} : \frac{1}{u} + \frac{1}{u ^{2}}

= - \int \frac{1}{u} + \frac{1}{u ^{2}} d u

Apply the Sum Rule:

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

= - (\int \frac{1}{u} d u + \int \frac{1}{u ^{2}} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

= - (ln ∣ u ∣ - \frac{1}{u})

Substitute back

u = cos (x)

= - (ln ∣ cos (x) ∣ - \frac{1}{cos ( x )})

Use the basic trigonometric identity:

\frac{1}{cos ( x )} = sec (x)

= - (ln ∣ cos (x) ∣ - sec (x))

Add a constant to the solution

= - (ln ∣ cos (x) ∣ - sec (x)) + C

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