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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int - \frac{1}{2 v} d v

= 2 \cdot \int - \frac{1}{2 v} d v

Take the constant out:

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

= 2 (- \frac{1}{2} \cdot \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 (- \frac{1}{2} ln ∣ v ∣)

Substitute back

= 2 (- \frac{1}{2} ln 1 - tan^{2} (\frac{x}{2}))

Simplify

2 (- \frac{1}{2} ln 1 - tan^{2} (\frac{x}{2})) : - ln 1 - tan^{2} (\frac{x}{2})

= - ln 1 - tan^{2} (\frac{x}{2})

Add a constant to the solution

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

6 - x^{2}, x^{2} - 2, [0, 3]

\int \frac{ln ( y )}{ln ( y )} d y

e^{t^{2}} y^{^{'}} + (t + t y^{2}) = 0, y (0) = 1

\frac{64 x}{x ^{2} + 64}

\frac{\partial}{\partial y} (2 x^{3} - x^{2} y)

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