What is the integral of 1/(3sin(x)+4cos(x)+5) ?

The integral of 1/(3sin(x)+4cos(x)+5) is -2/(tan(x/2)+3)+C

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integral of 1/(3sin(x)+4cos(x)+5)

\int \frac{1}{3 sin ( x ) + 4 cos ( x ) + 5} d x

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

Solution steps

\int \frac{1}{3 sin ( x ) + 4 cos ( x ) + 5} d x

Apply u-substitution

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

Take the partial fraction of

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

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

Apply u-substitution

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

Apply the Power Rule

= 2 (- \frac{1}{v})

Substitute back

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

Simplify

2 (- \frac{1}{tan ( \frac{x}{2} ) + 3}) : - \frac{2}{tan ( \frac{x}{2} ) + 3}

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

Add a constant to the solution

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

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What is the integral of 1/(3sin(x)+4cos(x)+5) ?
The integral of 1/(3sin(x)+4cos(x)+5) is -2/(tan(x/2)+3)+C