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

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

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

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

- (- sec (x) + ln ∣ cos (x) ∣) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

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

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

= - (- \frac{1}{u} + ln ∣ u ∣)

Substitute back

u = cos (x)

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

Use the basic trigonometric identity:

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

= - (- sec (x) + ln ∣ cos (x) ∣)

Add a constant to the solution

= - (- sec (x) + ln ∣ cos (x) ∣) + C

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