What is the integral of 1/(4-5sin(x)) ?

The integral of 1/(4-5sin(x)) is -1/3 ln|2tan(x/2)-1|+1/3 ln|tan(x/2)-2|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(4-5sin(x))

\int \frac{1}{4 - 5 sin ( x )} d x

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

Solution steps

\int \frac{1}{4 - 5 sin ( x )} d x

Apply u-substitution

= \int \frac{1}{2 ( u ^{2} + 1 ) - 5 u} d u

Take the partial fraction of

\frac{1}{2 ( u ^{2} + 1 ) - 5 u} : - \frac{2}{3 ( 2 u - 1 )} + \frac{1}{3 ( u - 2 )}

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

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

\int \frac{2}{3 ( 2 u - 1 )} d u = \frac{1}{3} ln ∣ 2 u - 1 ∣

\int \frac{1}{3 ( u - 2 )} d u = \frac{1}{3} ln ∣ u - 2 ∣

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

Substitute back

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

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

Add a constant to the solution

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

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