What is the integral of 1/(2sin(x)-cos(x)+3) ?

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

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integral of 1/(2sin(x)-cos(x)+3)

\int \frac{1}{2 sin ( x ) - cos ( x ) + 3} d x

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

Solution steps

\int \frac{1}{2 sin ( x ) - cos ( x ) + 3} d x

Apply u-substitution

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

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

Complete the square

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

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

Apply u-substitution

: \int \frac{2}{4 v ^{2} + 1} d v

= \int \frac{2}{4 v ^{2} + 1} d v

Take the constant out:

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

= 2 \cdot \int \frac{1}{4 v ^{2} + 1} d v

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= 2 \cdot \frac{1}{2} arctan (w)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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