What is the integral from 0 to 3 of 1/(2-cos(x)) ?

The integral from 0 to 3 of 1/(2-cos(x)) is 2/(sqrt(3))arctan(sqrt(3)tan(3/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 3 of 1/(2-cos(x))

\int_{0}^{3} \frac{1}{2 - cos ( x )} d x

\frac{2}{3} arctan (3 tan (\frac{3}{2}))

Decimal

1.76654 \dots

Solution steps

\int_{0}^{3} \frac{1}{2 - cos ( x )} d x

Apply u-substitution

: \int_{0}^{t a n (\frac{3}{2})} \frac{2}{3 u ^{2} + 1} d u

= \int_{0}^{t a n (\frac{3}{2})} \frac{2}{3 u ^{2} + 1} d u

Take the constant out:

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

= 2 \cdot \int_{0}^{t a n (\frac{3}{2})} \frac{1}{3 u ^{2} + 1} d u

Apply Integral Substitution

= 2 \cdot \int_{0}^{3 t a n (\frac{3}{2})} \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_{0}^{3 t a n (\frac{3}{2})} \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)]_{0}^{3 t a n (\frac{3}{2})}

Simplify

2 \cdot \frac{1}{3} [arctan (v)]_{0}^{3 t a n (\frac{3}{2})} : \frac{2}{3} [arctan (v)]_{0}^{3 t a n (\frac{3}{2})}

= \frac{2}{3} [arctan (v)]_{0}^{3 t a n (\frac{3}{2})}

Compute the boundaries:

arctan (3 tan (\frac{3}{2}))

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

\int_{0}^{a r c s i n (x)} cos (t) d t

\int_{0}^{2} y d y

\int_{- 1}^{1} x e^{x^{8}} d x

\int_{0}^{1} \frac{e ^{x}}{e ^{x} - 1} d x

\int_{0}^{\frac{π}{4}} cos^{2} (3 x) d x

What is the integral from 0 to 3 of 1/(2-cos(x)) ?
The integral from 0 to 3 of 1/(2-cos(x)) is 2/(sqrt(3))arctan(sqrt(3)tan(3/2))