What is the integral of (sin(4x))/(cos(4x)) ?

The integral of (sin(4x))/(cos(4x)) is -1/4 ln|cos(4x)|+C

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integral of (sin(4x))/(cos(4x))

\int \frac{sin ( 4 x )}{cos ( 4 x )} d x

- \frac{1}{4} ln ∣ cos (4 x) ∣ + C

Solution steps

\int \frac{sin ( 4 x )}{cos ( 4 x )} d x

Apply u-substitution

: \int - \frac{1}{4 u} d u

= \int - \frac{1}{4 u} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{4} ln ∣ u ∣

Substitute back

u = cos (4 x)

= - \frac{1}{4} ln ∣ cos (4 x) ∣

Add a constant to the solution

= - \frac{1}{4} ln ∣ cos (4 x) ∣ + C

f (x) = \frac{x ^{2}}{x - 5}

\frac{1}{s ( s + 1 ) ( s + 2 )}

\int \frac{cos ( x )}{1 + ( sin ( x ) ) ^{2}} d x

\int \frac{1}{x 25 x ^{2} + 49} d x

\frac{1}{2 s ^{2} + 6 s + 4}

What is the integral of (sin(4x))/(cos(4x)) ?
The integral of (sin(4x))/(cos(4x)) is -1/4 ln|cos(4x)|+C