What is the integral of 1/(5+4cos(θ)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(5+4cos(θ))

\int \frac{1}{5 + 4 cos ( θ )} d θ

\frac{2}{3} arctan (\frac{1}{3} tan (\frac{θ}{2})) + C

Solution steps

\int \frac{1}{5 + 4 cos ( θ )} d θ

Apply u-substitution

: \int \frac{2}{u ^{2} + 9} d u

= \int \frac{2}{u ^{2} + 9} 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}{u ^{2} + 9} d u

Apply Integral Substitution

= 2 \cdot \int \frac{1}{3 ( 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}{3} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 2 \cdot \frac{1}{3} arctan (v)

Substitute back

= 2 \cdot \frac{1}{3} arctan (\frac{tan ( \frac{θ}{2} )}{3})

Simplify

2 \cdot \frac{1}{3} arctan (\frac{tan ( \frac{θ}{2} )}{3}) : \frac{2}{3} arctan (\frac{1}{3} tan (\frac{θ}{2}))

= \frac{2}{3} arctan (\frac{1}{3} tan (\frac{θ}{2}))

Add a constant to the solution

= \frac{2}{3} arctan (\frac{1}{3} tan (\frac{θ}{2})) + C

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

\int (\frac{10 y + 11}{4 y ^{2} - 4 y + 5}) d y

\int 8 x (x^{2} + 1)^{3} d x

\int (2 x^{3} - \frac{1}{x ^{4}}) d x

\int \frac{e ^{3 x} + 5 e ^{2 x} - 4 e ^{x}}{e ^{x}} d x

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