What is the integral of 1/(2sin(x)-cos(x)+5) ?

The integral of 1/(2sin(x)-cos(x)+5) is 1/(sqrt(5))arctan(1/(sqrt(5))(3tan(x/2)+1))+C

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integral of 1/(2sin(x)-cos(x)+5)

\int \frac{1}{2 sin ( x ) - cos ( x ) + 5} d x

\frac{1}{5} arctan (\frac{1}{5} (3 tan (\frac{x}{2}) + 1)) + C

Solution steps

\int \frac{1}{2 sin ( x ) - cos ( x ) + 5} d x

Apply u-substitution

: \int \frac{1}{3 u ^{2} + 2 u + 2} d u

= \int \frac{1}{3 u ^{2} + 2 u + 2} d u

Complete the square

3 u^{2} + 2 u + 2 : 3 (u + \frac{1}{3})^{2} + \frac{5}{3}

= \int \frac{1}{3 ( u + \frac{1}{3} ) ^{2} + \frac{5}{3}} d u

Apply u-substitution

: \int \frac{3}{9 v ^{2} + 5} d v

= \int \frac{3}{9 v ^{2} + 5} d v

Take the constant out:

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

= 3 \cdot \int \frac{1}{9 v ^{2} + 5} d v

Apply Integral Substitution

= 3 \cdot \int \frac{1}{3 5 ( w ^{2} + 1 )} d w

Take the constant out:

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

= 3 \cdot \frac{1}{3 5} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= 3 \cdot \frac{1}{3 5} arctan (w)

Substitute back

= 3 \cdot \frac{1}{3 5} arctan (\frac{3}{5} (tan (\frac{x}{2}) + \frac{1}{3}))

Simplify

3 \cdot \frac{1}{3 5} arctan (\frac{3}{5} (tan (\frac{x}{2}) + \frac{1}{3})) : \frac{1}{5} arctan (\frac{1}{5} (3 tan (\frac{x}{2}) + 1))

= \frac{1}{5} arctan (\frac{1}{5} (3 tan (\frac{x}{2}) + 1))

Add a constant to the solution

= \frac{1}{5} arctan (\frac{1}{5} (3 tan (\frac{x}{2}) + 1)) + C

\frac{d}{d x} ((5)^{x})

\frac{\partial}{\partial x} (x^{n})

\frac{\partial}{\partial x} (x e^{y^{2}})

(2 t^{2} + 1)^{^{'}}

What is the integral of 1/(2sin(x)-cos(x)+5) ?
The integral of 1/(2sin(x)-cos(x)+5) is 1/(sqrt(5))arctan(1/(sqrt(5))(3tan(x/2)+1))+C