What is the integral of (cos(4x))/(sin(2x)) ?

The integral of (cos(4x))/(sin(2x)) is 1/2 (ln|tan(x)|+2cos(2x))+C

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integral of (cos(4x))/(sin(2x))

\int \frac{cos ( 4 x )}{sin ( 2 x )} d x

\frac{1}{2} (ln ∣ tan (x) ∣ + 2 cos (2 x)) + C

Solution steps

\int \frac{cos ( 4 x )}{sin ( 2 x )} d x

Rewrite using trig identities

= \int cos (4 x) csc (2 x) d x

Apply u-substitution

: \int \frac{1}{2} cos (2 u) csc (u) d u

= \int \frac{1}{2} cos (2 u) csc (u) d u

Take the constant out:

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

= \frac{1}{2} \cdot \int cos (2 u) csc (u) d u

Rewrite using trig identities

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

Expand

\frac{1 - 2 sin ^{2} ( u )}{sin ( u )} : \frac{1}{sin ( u )} - 2 sin (u)

= \frac{1}{2} \cdot \int \frac{1}{sin ( u )} - 2 sin (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

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

\int \frac{1}{sin ( u )} d u = ln tan (\frac{u}{2})

\int 2 sin (u) d u = - 2 cos (u)

= \frac{1}{2} (ln tan (\frac{u}{2}) - (- 2 cos (u)))

Substitute back

u = 2 x

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

Simplify

\frac{1}{2} (ln tan (\frac{2 x}{2}) - (- 2 cos (2 x))) : \frac{1}{2} (ln ∣ tan (x) ∣ + 2 cos (2 x))

= \frac{1}{2} (ln ∣ tan (x) ∣ + 2 cos (2 x))

Add a constant to the solution

= \frac{1}{2} (ln ∣ tan (x) ∣ + 2 cos (2 x)) + C

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What is the integral of (cos(4x))/(sin(2x)) ?
The integral of (cos(4x))/(sin(2x)) is 1/2 (ln|tan(x)|+2cos(2x))+C