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

The integral of arctan((sin(x))/(1+cos(x))) is arctan^2(tan(x/2))+C

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

= 2 \cdot \int \frac{arctan ( u )}{1 + u ^{2}} d u

Apply u-substitution

: \int v d v

= 2 \cdot \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= 2 \cdot \frac{v ^{2}}{2}

Substitute back

= 2 \cdot \frac{arctan ^{2} ( tan ( \frac{x}{2} ) )}{2}

Simplify

2 \cdot \frac{arctan ^{2} ( tan ( \frac{x}{2} ) )}{2} : arctan^{2} (tan (\frac{x}{2}))

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

Add a constant to the solution

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

(\frac{x ^{2} + 4}{x ^{2} - 4})^{4}

\frac{d y}{d x} = (y - 1)^{3}

y = \frac{4 x + 5}{x - 1}, (2, 13)

\frac{d}{d x} (a sec (x))

\int \frac{ln ( x ^{6} )}{x} d x

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