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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{2}{u ^{2} + 4 u + 5} d u

= \int \frac{2}{u ^{2} + 4 u + 5} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{u ^{2} + 4 u + 5} d u

Complete the square

u^{2} + 4 u + 5 : (u + 2)^{2} + 1

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

Apply u-substitution

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

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

Use the common integral:

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

= 2 arctan (v)

Substitute back

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

Add a constant to the solution

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

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