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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

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

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

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

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

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

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

Substitute back

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

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

Add a constant to the solution

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

\frac{d}{d x} (sin (e^{x^{3} s i n (x)}))

\frac{d}{d x} (x^{2} - 8 x + 12)

\int \frac{1}{8 - 3 x} d x

\int_{0}^{1} - ln (x) d x

\frac{d y}{d x} = - y (x + 1)

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