What is the integral of 1/(5+3cos(2x)) ?

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

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integral of 1/(5+3cos(2x))

\int \frac{1}{5 + 3 cos ( 2 x )} d x

\frac{1}{4} arctan (\frac{tan ( x )}{2}) + C

Solution steps

\int \frac{1}{5 + 3 cos ( 2 x )} d x

Apply u-substitution

: \int \frac{1}{2 ( 5 + 3 cos ( u ) )} d u

= \int \frac{1}{2 ( 5 + 3 cos ( u ) )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{5 + 3 cos ( u )} d u

Apply u-substitution

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{1}{2} arctan (\frac{tan ( \frac{2 x}{2} )}{2}) : \frac{1}{4} arctan (\frac{tan ( x )}{2})

= \frac{1}{4} arctan (\frac{tan ( x )}{2})

Add a constant to the solution

= \frac{1}{4} arctan (\frac{tan ( x )}{2}) + C

\int 7 x^{3} e^{- x^{2}} d x

\frac{\partial}{\partial y} ((y - 2) (2 x + y - 5))

e^{x} y \frac{d y}{d x} = e^{- y} + e^{- 4 x - y}

\int_{0}^{1} (x^{2} - 3 x + 5) d x

f (x) = \frac{2}{3 x ^{\frac{2}{3}}}, at x = 64

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