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

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

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

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

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

Solution steps

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

Expand

\frac{x + sin ( x )}{1 + cos ( x )} : \frac{x}{1 + cos ( x )} + \frac{sin ( x )}{1 + cos ( x )}

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

\int \frac{x}{1 + cos ( x )} d x = x tan (\frac{x}{2}) + 2 ln cos (\frac{x}{2})

\int \frac{sin ( x )}{1 + cos ( x )} d x = - ln ∣ 1 + cos (x) ∣

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

Add a constant to the solution

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

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = - 7, y^{^{'}} (0) = 28

(2 e^{2 x})^{^{'}}

- 9 (6 x^{2} + 7)^{- 6}

(x^{2} + y^{2})^{2}

\frac{\partial}{\partial y} (3 z e^{x yz})

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