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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integral Substitution

= \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

= \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)

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

Substitute back

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

Add a constant to the solution

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

\int_{- π}^{π} ∣ sin (x) ∣ d x

y^{^{'}} = (3 x - 2) (30 x + 19)

\frac{\partial}{\partial x} (400 \cdot sin (x) \cdot cos (x))

\frac{\partial}{\partial x} (y^{2} e^{x y^{2} + z^{2}})

y^{^{'}} = x^{2} e^{2 y}, y (- 1) = - 1

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