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

Solution

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

\frac{1}{6} (ln \frac{tan ( \frac{x}{2} ) - 1}{6} + 1 - ln \frac{tan ( \frac{x}{2} ) - 1}{6} - 1) + C

Solution steps

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

Apply u-substitution

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

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

Complete the square

- u^{2} + 2 u + 5 : - (u - 1)^{2} + 6

= 2 \cdot \int \frac{1}{- ( u - 1 ) ^{2} + 6} d u

Apply u-substitution

: \int \frac{1}{- v ^{2} + 6} d v

= 2 \cdot \int \frac{1}{- v ^{2} + 6} d v

Apply Integral Substitution

= 2 \cdot \int \frac{1}{6 ( - 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}{6} \cdot \int \frac{1}{- w ^{2} + 1} d w

Use the common integral:

\int \frac{1}{- w ^{2} + 1} d w = \frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}

= 2 \cdot \frac{1}{6} (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2})

Substitute back

= 2 \cdot \frac{1}{6} \frac{ln \frac{t a n ( \frac{x}{2} ) - 1}{6} + 1}{2} - \frac{ln \frac{t a n ( \frac{x}{2} ) - 1}{6} - 1}{2}

Simplify

2 \cdot \frac{1}{6} \frac{ln \frac{t a n ( \frac{x}{2} ) - 1}{6} + 1}{2} - \frac{ln \frac{t a n ( \frac{x}{2} ) - 1}{6} - 1}{2} : \frac{1}{6} (ln \frac{tan ( \frac{x}{2} ) - 1}{6} + 1 - ln \frac{tan ( \frac{x}{2} ) - 1}{6} - 1)

= \frac{1}{6} (ln \frac{tan ( \frac{x}{2} ) - 1}{6} + 1 - ln \frac{tan ( \frac{x}{2} ) - 1}{6} - 1)

Add a constant to the solution

= \frac{1}{6} (ln \frac{tan ( \frac{x}{2} ) - 1}{6} + 1 - ln \frac{tan ( \frac{x}{2} ) - 1}{6} - 1) + C

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