What is the integral of (tan(x))/(sec(x)+4cos(x)) ?

The integral of (tan(x))/(sec(x)+4cos(x)) is -1/2 arctan(2cos(x))+C

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integral of (tan(x))/(sec(x)+4cos(x))

\int \frac{tan ( x )}{sec ( x ) + 4 cos ( x )} d x

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

Solution steps

\int \frac{tan ( x )}{sec ( x ) + 4 cos ( x )} d x

Apply Trigonometric Substitution

: \int - \frac{1}{1 + 4 u ^{2}} d u

= \int - \frac{1}{1 + 4 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}{1 + 4 u ^{2}} d u

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{1 + v ^{2}} d v = arctan (v)

= - \frac{1}{2} arctan (v)

Substitute back

= - \frac{1}{2} arctan (2 cos (x))

Add a constant to the solution

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

\int 12 x + 5 d x

\frac{\partial}{\partial x} (cos (4 y^{2} - 2 x))

\frac{d}{d x} (sin (x^{t a n (x)}))

(2 x + 1) y^{^{'}} + 2 y = 4 x, y (0) = 1

\frac{\partial}{\partial z} (ln (x^{2} + y^{4}) - z)

What is the integral of (tan(x))/(sec(x)+4cos(x)) ?
The integral of (tan(x))/(sec(x)+4cos(x)) is -1/2 arctan(2cos(x))+C