What is the integral from pi/6 to pi/3 of 1/(sin(x)) ?

The integral from pi/6 to pi/3 of 1/(sin(x)) is -1/2 ln(3)-ln(2-sqrt(3))

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integral from pi/6 to pi/3 of 1/(sin(x))

\int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{sin ( x )} d x

- \frac{1}{2} ln (3) - ln (2 - 3)

Solution steps

\int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{sin ( x )} d x

Apply u-substitution

: \int_{2 - 3}^{\frac{3}{3}} \frac{1}{u} d u

= \int_{2 - 3}^{\frac{3}{3}} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{2 - 3}^{\frac{3}{3}}

\frac{3}{3} = \frac{1}{3}

= [ln ∣ u ∣]_{2 - 3}^{\frac{1}{3}}

Compute the boundaries:

- \frac{1}{2} ln (3) - ln (2 - 3)

= - \frac{1}{2} ln (3) - ln (2 - 3)

\int_{3}^{4} x - 3 d x

\int_{- 1}^{2} x 2 + x d x

\int_{x}^{2 x} (x^{2} + y^{2}) d y

\int_{0}^{π} - sin^{2} (x) d x

\int_{0}^{1} 1 e^{- s t} d t

What is the integral from pi/6 to pi/3 of 1/(sin(x)) ?
The integral from pi/6 to pi/3 of 1/(sin(x)) is -1/2 ln(3)-ln(2-sqrt(3))