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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{2}{u ^{2} + 4 u + 3} d u

= \int \frac{2}{u ^{2} + 4 u + 3} 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} + 4 u + 3} d u

Take the partial fraction of

\frac{1}{u ^{2} + 4 u + 3} : \frac{1}{2 ( u + 1 )} - \frac{1}{2 ( u + 3 )}

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

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

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

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

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{x ^{2}}{x ^{4} + 2 x ^{2} + 1} d x

\int \frac{1}{x ( ln ( x ^{2} ) ) ^{7}} d x

\int \frac{x ^{5}}{1 - x ^{3}} d x

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

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

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