What is the integral of 1/(3cos(u)-5-4sin(u)) ?

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

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integral of 1/(3cos(u)-5-4sin(u))

\int \frac{1}{3 cos ( u ) - 5 - 4 sin ( u )} d u

\frac{1}{2 ( 2 tan ( \frac{u}{2} ) + 1 )} + C

Solution steps

\int \frac{1}{3 cos ( u ) - 5 - 4 sin ( u )} d u

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Complete the square

4 v^{2} + 4 v + 1 : 4 (v + \frac{1}{2})^{2}

= - \int \frac{1}{4 ( v + \frac{1}{2} ) ^{2}} d v

Apply u-substitution

: \int \frac{1}{4 w ^{2}} d w

= - \int \frac{1}{4 w ^{2}} d w

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{4} \cdot \int \frac{1}{w ^{2}} d w

Apply the Power Rule

: - \frac{1}{w}

= - \frac{1}{4} (- \frac{1}{w})

Substitute back

= - \frac{1}{4} (- \frac{1}{tan ( \frac{u}{2} ) + \frac{1}{2}})

Simplify

- \frac{1}{4} (- \frac{1}{tan ( \frac{u}{2} ) + \frac{1}{2}}) : \frac{1}{2 ( 2 tan ( \frac{u}{2} ) + 1 )}

= \frac{1}{2 ( 2 tan ( \frac{u}{2} ) + 1 )}

Add a constant to the solution

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

\int_{0}^{1} arcsin (x) d x

\frac{d}{d x} (x^{4 π})

\frac{\partial}{\partial x} (y^{\frac{1}{3}} x^{3})

\frac{d}{d v} (\frac{u}{v})

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