What is the integral of 1/(1-sin(1/2)x) ?

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

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integral of 1/(1-sin(1/2)x)

\int \frac{1}{1 - sin ( \frac{1}{2} ) x} d x

- csc (\frac{1}{2}) ln 1 - sin (\frac{1}{2}) x + C

Solution steps

\int \frac{1}{1 - sin ( \frac{1}{2} ) x} d x

Apply u-substitution

: \int - \frac{1}{sin ( \frac{1}{2} ) u} d u

= \int - \frac{1}{sin ( \frac{1}{2} ) u} d u

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{sin ( \frac{1}{2} )} ln ∣ u ∣

Substitute back

u = 1 - sin (\frac{1}{2}) x

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

Use the basic trigonometric identity:

\frac{1}{sin ( x )} = csc (x)

= - csc (\frac{1}{2}) ln 1 - sin (\frac{1}{2}) x

Add a constant to the solution

= - csc (\frac{1}{2}) ln 1 - sin (\frac{1}{2}) x + C

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