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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Complete the square

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

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

= - \frac{1}{v}

Substitute back

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

Add a constant to the solution

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

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