What is the integral of 1/(4sin(x)) ?

The integral of 1/(4sin(x)) is 1/4 ln|tan(x/2)|+C

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integral of 1/(4sin(x))

\int \frac{1}{4 sin ( x )} d x

\frac{1}{4} ln tan (\frac{x}{2}) + C

Solution steps

\int \frac{1}{4 sin ( x )} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{sin ( x )} d x

Apply u-substitution

: \int \frac{1}{u} d u

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = tan (\frac{x}{2})

= \frac{1}{4} ln tan (\frac{x}{2})

Add a constant to the solution

= \frac{1}{4} ln tan (\frac{x}{2}) + C

x \to 0 lim (lo g_{\frac{2}{3}} (x))

\frac{\partial}{\partial x} (5 e^{x y + 6})

y^{^{''}} + 2 y^{^{'}} + y = 6 e^{- t}

\int \frac{1}{4} e^{- 2 x} d x

f (x) = 4 x^{2} - 5 x + 2

What is the integral of 1/(4sin(x)) ?
The integral of 1/(4sin(x)) is 1/4 ln|tan(x/2)|+C