What is the integral of 1/(2sin(θ)) ?

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

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integral of 1/(2sin(θ))

\int \frac{1}{2 sin ( θ )} d θ

\frac{1}{2} ln tan (\frac{θ}{2}) + C

Solution steps

\int \frac{1}{2 sin ( θ )} d θ

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{sin ( θ )} d θ

Apply u-substitution

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = tan (\frac{θ}{2})

= \frac{1}{2} ln tan (\frac{θ}{2})

Add a constant to the solution

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

\frac{\partial ^{2}}{\partial u ( v ) \partial v} (8 v e^{- 6 u (v)})

\int \frac{2 cos ( x )}{- 2 + 6 sin ( x )} d x

\frac{16 x}{x ^{2} + 16}

7 x^{2} e^{x}

f (x) = - 4 (2 x + 7)^{- 5} (2), at x = 2

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