What is the integral of 1/(3sin(x)-2cos(x)+7) ?

The integral of 1/(3sin(x)-2cos(x)+7) is 1/3 arctan(3/2 tan(x/2)+1/2)+C

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integral of 1/(3sin(x)-2cos(x)+7)

\int \frac{1}{3 sin ( x ) - 2 cos ( x ) + 7} d x

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

Solution steps

\int \frac{1}{3 sin ( x ) - 2 cos ( x ) + 7} d x

Apply u-substitution

: \int \frac{2}{9 u ^{2} + 6 u + 5} d u

= \int \frac{2}{9 u ^{2} + 6 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}{9 u ^{2} + 6 u + 5} d u

Complete the square

9 u^{2} + 6 u + 5 : 9 (u + \frac{1}{3})^{2} + 4

= 2 \cdot \int \frac{1}{9 ( u + \frac{1}{3} ) ^{2} + 4} d u

Apply u-substitution

: \int \frac{1}{9 v ^{2} + 4} d v

= 2 \cdot \int \frac{1}{9 v ^{2} + 4} 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 = arctan (w)

= 2 \cdot \frac{1}{6} arctan (w)

Substitute back

= 2 \cdot \frac{1}{6} arctan (\frac{3}{2} (tan (\frac{x}{2}) + \frac{1}{3}))

Simplify

2 \cdot \frac{1}{6} arctan (\frac{3}{2} (tan (\frac{x}{2}) + \frac{1}{3})) : \frac{1}{3} arctan (\frac{3}{2} tan (\frac{x}{2}) + \frac{1}{2})

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

Add a constant to the solution

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

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What is the integral of 1/(3sin(x)-2cos(x)+7) ?
The integral of 1/(3sin(x)-2cos(x)+7) is 1/3 arctan(3/2 tan(x/2)+1/2)+C