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

The integral of 1/(3+3cos(x)) is 1/3 tan(x/2)+C

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

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

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

Solution steps

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

Factor

\frac{1}{3 + 3 cos ( x )}

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

Take the constant out:

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

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

Apply u-substitution

= \frac{1}{3} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot u

Substitute back

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

= \frac{1}{3} \cdot 1 \cdot tan (\frac{x}{2})

Simplify

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (ln (x cos (y)))

d er i v a t i v e \frac{5}{5 t} + 5 t

\frac{d}{d x} ((x)^{3})

\int_{1}^{2} 8 r^{4} ln (r) d r

y^{^{'''}} + y^{^{''}} - 2 = 0

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