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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

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}{1 + cos ( u )} d u

Apply u-substitution

: \int 1 d v

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

Integral of a constant:

\int a d x = a x

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

Substitute back

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

Simplify

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (22 x y e^{- x^{2} - y^{2}})

- 0.1 x^{2} + 1.2 x + 98.4

2 x y d x + (x^{2} + 2 y) d y = 0

y = tan^{2} (x)

\int \frac{1}{sin ( t )} d t

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