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

The integral of 1/(2+sin(θ)+cos(θ)) is sqrt(2)arctan(1/(sqrt(2))(tan(θ/2)+1))+C

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integral of 1/(2+sin(θ)+cos(θ))

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

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

Solution steps

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

Apply u-substitution

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

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

Complete the square

u^{2} + 2 u + 3 : (u + 1)^{2} + 2

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

Apply u-substitution

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

= 2 \cdot \int \frac{1}{v ^{2} + 2} 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 (\frac{tan ( \frac{θ}{2} ) + 1}{2})

Simplify

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

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

Add a constant to the solution

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

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