What is the integral of 1/(3+2cos(x)+3sin(x)) ?

The integral of 1/(3+2cos(x)+3sin(x)) is 2(1/4 ln|tan(x/2)+1|-1/4 ln|tan(x/2)+5|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(3+2cos(x)+3sin(x))

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

2 (\frac{1}{4} ln tan (\frac{x}{2}) + 1 - \frac{1}{4} ln tan (\frac{x}{2}) + 5) + C

Solution steps

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

Apply u-substitution

: \int \frac{2}{u ^{2} + 6 u + 5} d u

= \int \frac{2}{u ^{2} + 6 u + 5} 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} + 6 u + 5} d u

Take the partial fraction of

\frac{1}{u ^{2} + 6 u + 5} : \frac{1}{4 ( u + 1 )} - \frac{1}{4 ( u + 5 )}

= 2 \cdot \int \frac{1}{4 ( u + 1 )} - \frac{1}{4 ( u + 5 )} 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}{4 ( u + 1 )} d u - \int \frac{1}{4 ( u + 5 )} d u)

\int \frac{1}{4 ( u + 1 )} d u = \frac{1}{4} ln ∣ u + 1 ∣

\int \frac{1}{4 ( u + 5 )} d u = \frac{1}{4} ln ∣ u + 5 ∣

= 2 (\frac{1}{4} ln ∣ u + 1 ∣ - \frac{1}{4} ln ∣ u + 5 ∣)

Substitute back

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

= 2 (\frac{1}{4} ln tan (\frac{x}{2}) + 1 - \frac{1}{4} ln tan (\frac{x}{2}) + 5)

Add a constant to the solution

= 2 (\frac{1}{4} ln tan (\frac{x}{2}) + 1 - \frac{1}{4} ln tan (\frac{x}{2}) + 5) + C

arccos (2 x)

f (x) = x + 6, at x = 1

\frac{x ^{2}}{7 + 6 x}

\frac{d}{d x} (R (x) (x)^{2} - x^{2})

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

What is the integral of 1/(3+2cos(x)+3sin(x)) ?
The integral of 1/(3+2cos(x)+3sin(x)) is 2(1/4 ln|tan(x/2)+1|-1/4 ln|tan(x/2)+5|)+C