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

The integral of 1/(2+5cos(x)) is 1/(sqrt(21))(ln|sqrt(3/7)tan(x/2)+1|-ln|sqrt(3/7)tan(x/2)-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{21} (ln \frac{3}{7} tan (\frac{x}{2}) + 1 - ln \frac{3}{7} tan (\frac{x}{2}) - 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{2}{- 3 u ^{2} + 7} d u

= \int \frac{2}{- 3 u ^{2} + 7} 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 u ^{2} + 7} d u

Apply Integral Substitution

= 2 \cdot \int \frac{1}{21 ( - v ^{2} + 1 )} d v

Take the constant out:

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

= 2 \cdot \frac{1}{21} \cdot \int \frac{1}{- v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 2 \cdot \frac{1}{21} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= 2 \cdot \frac{1}{21} \frac{ln \frac{3}{7} tan ( \frac{x}{2} ) + 1}{2} - \frac{ln \frac{3}{7} tan ( \frac{x}{2} ) - 1}{2}

Simplify

2 \cdot \frac{1}{21} \frac{ln \frac{3}{7} tan ( \frac{x}{2} ) + 1}{2} - \frac{ln \frac{3}{7} tan ( \frac{x}{2} ) - 1}{2} : \frac{1}{21} (ln \frac{3}{7} tan (\frac{x}{2}) + 1 - ln \frac{3}{7} tan (\frac{x}{2}) - 1)

= \frac{1}{21} (ln \frac{3}{7} tan (\frac{x}{2}) + 1 - ln \frac{3}{7} tan (\frac{x}{2}) - 1)

Add a constant to the solution

= \frac{1}{21} (ln \frac{3}{7} tan (\frac{x}{2}) + 1 - ln \frac{3}{7} tan (\frac{x}{2}) - 1) + C

\int x (x^{2} + 1) d x

\int (5 x - 3)^{3} d x

\int t sin (t) d t

\int - 2 t^{4} e^{6 t^{5}} d t

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

What is the integral of 1/(2+5cos(x)) ?
The integral of 1/(2+5cos(x)) is 1/(sqrt(21))(ln|sqrt(3/7)tan(x/2)+1|-ln|sqrt(3/7)tan(x/2)-1|)+C