What is the integral of (cos(4x)+1)/(2sin(2x)) ?

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

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integral of (cos(4x)+1)/(2sin(2x))

integral

\frac{cos ( 4 x ) + 1}{2 sin ( 2 x )}

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

Solution steps

\int \frac{cos ( 4 x ) + 1}{2 sin ( 2 x )} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{cos ( 4 x ) + 1}{sin ( 2 x )} d x

Expand

\frac{cos ( 4 x ) + 1}{sin ( 2 x )} : \frac{cos ( 4 x )}{sin ( 2 x )} + \frac{1}{sin ( 2 x )}

= \frac{1}{2} \cdot \int \frac{cos ( 4 x )}{sin ( 2 x )} + \frac{1}{sin ( 2 x )} d x

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{cos ( 4 x )}{sin ( 2 x )} d x + \int \frac{1}{sin ( 2 x )} d x)

\int \frac{cos ( 4 x )}{sin ( 2 x )} d x = \frac{1}{2} (ln ∣ tan (x) ∣ + 2 cos (2 x))

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{4 x}{( x + 2 )}

(e^{x} - 1)^{2}

x + 1 x

\frac{x ^{2} - 81}{x ^{4}}

\frac{x ^{2} - 1}{x ^{2}}

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