What is the integral of 1/(5+13sin(2x)) ?

The integral of 1/(5+13sin(2x)) is 1/24 ln|5tan(x)+1|-1/24 ln|tan(x)+5|+C

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integral of 1/(5+13sin(2x))

\int \frac{1}{5 + 13 sin ( 2 x )} d x

\frac{1}{24} ln ∣ 5 tan (x) + 1 ∣ - \frac{1}{24} ln ∣ tan (x) + 5 ∣ + C

Solution steps

\int \frac{1}{5 + 13 sin ( 2 x )} d x

Apply u-substitution

: \int \frac{1}{2 ( 5 + 13 sin ( u ) )} d u

= \int \frac{1}{2 ( 5 + 13 sin ( 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 + 13 sin ( u )} d u

Apply u-substitution

: \int \frac{2}{5 ( 1 + v ^{2} ) + 26 v} d v

= \frac{1}{2} \cdot \int \frac{2}{5 ( 1 + v ^{2} ) + 26 v} d v

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{5 ( 1 + v ^{2} ) + 26 v} d v

Take the partial fraction of

\frac{1}{5 ( 1 + v ^{2} ) + 26 v} : \frac{5}{24 ( 5 v + 1 )} - \frac{1}{24 ( v + 5 )}

= \frac{1}{2} \cdot 2 \cdot \int \frac{5}{24 ( 5 v + 1 )} - \frac{1}{24 ( v + 5 )} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} \cdot 2 (\int \frac{5}{24 ( 5 v + 1 )} d v - \int \frac{1}{24 ( v + 5 )} d v)

\int \frac{5}{24 ( 5 v + 1 )} d v = \frac{1}{24} ln ∣ 5 v + 1 ∣

\int \frac{1}{24 ( v + 5 )} d v = \frac{1}{24} ln ∣ v + 5 ∣

= \frac{1}{2} \cdot 2 (\frac{1}{24} ln ∣ 5 v + 1 ∣ - \frac{1}{24} ln ∣ v + 5 ∣)

Substitute back

= \frac{1}{2} \cdot 2 (\frac{1}{24} ln 5 tan (\frac{2 x}{2}) + 1 - \frac{1}{24} ln tan (\frac{2 x}{2}) + 5)

Simplify

\frac{1}{2} \cdot 2 (\frac{1}{24} ln 5 tan (\frac{2 x}{2}) + 1 - \frac{1}{24} ln tan (\frac{2 x}{2}) + 5) : \frac{1}{24} ln ∣ 5 tan (x) + 1 ∣ - \frac{1}{24} ln ∣ tan (x) + 5 ∣

= \frac{1}{24} ln ∣ 5 tan (x) + 1 ∣ - \frac{1}{24} ln ∣ tan (x) + 5 ∣

Add a constant to the solution

= \frac{1}{24} ln ∣ 5 tan (x) + 1 ∣ - \frac{1}{24} ln ∣ tan (x) + 5 ∣ + C

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What is the integral of 1/(5+13sin(2x)) ?
The integral of 1/(5+13sin(2x)) is 1/24 ln|5tan(x)+1|-1/24 ln|tan(x)+5|+C