What is the integral of 5/(3-2cos(x)) ?

The integral of 5/(3-2cos(x)) is 2sqrt(5)arctan(sqrt(5)tan(x/2))+C

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integral of 5/(3-2cos(x))

\int \frac{5}{3 - 2 cos ( x )} d x

25 arctan (5 tan (\frac{x}{2})) + C

Solution steps

\int \frac{5}{3 - 2 cos ( x )} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{3 - 2 cos ( x )} d x

Apply u-substitution

: \int \frac{2}{5 u ^{2} + 1} d u

= 5 \cdot \int \frac{2}{5 u ^{2} + 1} d u

Take the constant out:

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

= 5 \cdot 2 \cdot \int \frac{1}{5 u ^{2} + 1} d u

Apply Integral Substitution

= 5 \cdot 2 \cdot \int \frac{1}{5 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 5 \cdot 2 \cdot \frac{1}{5} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 5 \cdot 2 \cdot \frac{1}{5} arctan (v)

Substitute back

= 5 \cdot 2 \cdot \frac{1}{5} arctan (5 tan (\frac{x}{2}))

Simplify

5 \cdot 2 \cdot \frac{1}{5} arctan (5 tan (\frac{x}{2})) : 25 arctan (5 tan (\frac{x}{2}))

= 25 arctan (5 tan (\frac{x}{2}))

Add a constant to the solution

= 25 arctan (5 tan (\frac{x}{2})) + C

x \to 3 lim (4 π)

\frac{\partial}{\partial x} (x cos (\frac{y}{z}))

\frac{d}{d x} (x + e^{- x})

\frac{\partial}{\partial y} (3 x^{2} y^{4})

\frac{d}{d x} (\frac{x - 2}{x - 2})

What is the integral of 5/(3-2cos(x)) ?
The integral of 5/(3-2cos(x)) is 2sqrt(5)arctan(sqrt(5)tan(x/2))+C