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

The integral of 1/(1+cos(x)+sin^2(x/2)) is sqrt(2)arctan(1/(sqrt(2))tan(x/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/(1+cos(x)+sin^2(x/2))

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

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

Solution steps

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

Rewrite using trig identities

= \int \frac{2}{cos ( x ) + 3} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{cos ( x ) + 3} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 2} d u

= 2 \cdot \int \frac{1}{u ^{2} + 2} d u

Apply Integral Substitution

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

Substitute back

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

Simplify

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

= 2 arctan (\frac{1}{2} tan (\frac{x}{2}))

Add a constant to the solution

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

\int \frac{sin ^{2} ( z )}{2} - sin (z) d z

\int (3 x - 1) ln (x) d x

\int \frac{4}{( x ^{2} + 2 )} d x

\int 36 x^{2} + 36 x^{4} d x

What is the integral of 1/(1+cos(x)+sin^2(x/2)) ?
The integral of 1/(1+cos(x)+sin^2(x/2)) is sqrt(2)arctan(1/(sqrt(2))tan(x/2))+C