# Evaluate $\displaystyle \int \dfrac{1+\cos{4x}}{\cot{x}-\tan{x}} dx$

$x$ is an angle. Trigonometric functions cos, tan and cot functions formed a trigonometric expression and $dx$ is an element of integration.

### Transform tan and cot functions

The denominator contains tan and cot functions. They have to be changed to another form to simplify the expression. The quotient trigonometric identities are useful to express tan and cot function in terms of sine and cosine functions.

$= \displaystyle \int \dfrac{1+\cos{4x}}{\dfrac{\cos{x}}{\sin{x}}-\dfrac{\sin{x}}{\cos{x}}} dx$

Now, simplify the expression to its maximum level.

$= \displaystyle \int \dfrac{1+\cos{4x}}{\dfrac{ \cos{x} \times \cos{x} -\sin{x} \times \sin{x}}{\sin{x} \times \cos{x}}} dx$

$= \displaystyle \int \dfrac{1+\cos{4x}}{\dfrac{\cos^2{x}-\sin^2{x}}{\sin{x}\cos{x}}} dx$

$= \displaystyle \int {[1+\cos{4x}]} \times \dfrac{\sin{x}\cos{x}}{\cos^2{x}-\sin^2{x}} dx$

$= \displaystyle \int \dfrac{[1+\cos{4x}]\sin{x}\cos{x}}{\cos^2{x}-\sin^2{x}} dx$

### Simplify the denominator trigonometric expression

The subtraction of square of sin of angle from square of cosine of angle represents cos double angle identity.

$= \displaystyle \int \dfrac{[1+\cos{4x}]\sin{x}\cos{x}}{\cos{2x}} dx$

### Simplify the numerator trigonometric expression

The numerator can be simplified by applying sum of one and cos of double angle trigonometric identity. In order to use this rule, a small mathematical adjustment should be made firstly.

$= \displaystyle \int \dfrac{[1+\cos{2(2x)}]\sin{x}\cos{x}}{\cos{2x}} dx$

As per sum of one and cosine of double angle trigonometric identity, $1+\cos{2\theta} = 2\cos^2{\theta}$

$= \displaystyle \int \dfrac{2\cos^2{2x} \times \sin{x}\cos{x}}{\cos{2x}} dx$

$= \displaystyle \int \require{cancel} \dfrac{\cancel{\cos^2{2x}} \times 2\sin{x}\cos{x}}{\cancel{\cos{2x}}} dx$

### Simplify the numerator trigonometric expression

The trigonometric expression can be simplified further by sine double angle identity.

$= \displaystyle \int \cos{2x} \times 2\sin{x}\cos{x}dx$

$= \displaystyle \int \cos{2x} \times \sin{2x} dx$

Sine double angle identity can also be used one more time.

$= \displaystyle \int \sin{2x}\cos{2x} dx$

$= \displaystyle \int \dfrac{1}{2} \times 2 \times \sin{2x}\cos{2x} dx$

$= \displaystyle \dfrac{1}{2} \int 2\sin{2x}\cos{2x} dx$

$= \displaystyle \dfrac{1}{2} \int \sin{2(2x)} dx$

$= \displaystyle \dfrac{1}{2} \int \sin{4x} dx$

### Find integration of the function

The trigonometric expression is successfully simplified and the integration of the function can be performed immediately.

The integration of sine function can be performed by the integral of sin function formula.

$\displaystyle \int \sin{x} dx = -\cos{x}+C$

This formula can be used directly due to different angle and element of integration. So, use replacement technique to make the integration of the function possible.

Take $u = 4x$, then $du = 4dx$. Therefore, $dx = \dfrac{du}{4}$

$= \displaystyle \dfrac{1}{2} \int \sin{u} \dfrac{du}{4}$

$= \displaystyle \dfrac{1}{2} \times \dfrac{1}{4} \int \sin{u} du$

$= \displaystyle \dfrac{1 \times 1}{2 \times 4} \int \sin{u} du$

$= \displaystyle \dfrac{1}{8} \int \sin{u} du$

$= \displaystyle \dfrac{1}{8} [-\cos{u}+C]$

The trigonometric function is actually in terms of $x$ but it is now in terms of $u$. So, replace $u$ by its value.

$= \displaystyle \dfrac{-\cos{4x}}{8}+\dfrac{C}{8}$

$= \displaystyle \dfrac{-\cos{4x}}{8}+C$

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