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

The integral of 1/(3+5sin(x)) is 2(1/8 ln|3tan(x/2)+1|-1/8 ln|tan(x/2)+3|)+C

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

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

2 (\frac{1}{8} ln 3 tan (\frac{x}{2}) + 1 - \frac{1}{8} ln tan (\frac{x}{2}) + 3) + C

Solution steps

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

Apply u-substitution

: \int \frac{2}{3 ( 1 + u ^{2} ) + 10 u} d u

= \int \frac{2}{3 ( 1 + u ^{2} ) + 10 u} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{3 ( 1 + u ^{2} ) + 10 u} d u

Take the partial fraction of

\frac{1}{3 ( 1 + u ^{2} ) + 10 u} : \frac{3}{8 ( 3 u + 1 )} - \frac{1}{8 ( u + 3 )}

= 2 \cdot \int \frac{3}{8 ( 3 u + 1 )} - \frac{1}{8 ( u + 3 )} d u

Apply the Sum Rule:

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

= 2 (\int \frac{3}{8 ( 3 u + 1 )} d u - \int \frac{1}{8 ( u + 3 )} d u)

\int \frac{3}{8 ( 3 u + 1 )} d u = \frac{1}{8} ln ∣ 3 u + 1 ∣

\int \frac{1}{8 ( u + 3 )} d u = \frac{1}{8} ln ∣ u + 3 ∣

= 2 (\frac{1}{8} ln ∣ 3 u + 1 ∣ - \frac{1}{8} ln ∣ u + 3 ∣)

Substitute back

u = tan (\frac{x}{2})

= 2 (\frac{1}{8} ln 3 tan (\frac{x}{2}) + 1 - \frac{1}{8} ln tan (\frac{x}{2}) + 3)

Add a constant to the solution

= 2 (\frac{1}{8} ln 3 tan (\frac{x}{2}) + 1 - \frac{1}{8} ln tan (\frac{x}{2}) + 3) + C

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What is the integral of 1/(3+5sin(x)) ?
The integral of 1/(3+5sin(x)) is 2(1/8 ln|3tan(x/2)+1|-1/8 ln|tan(x/2)+3|)+C