What is the integral of [sin(pi/3-x)]^{-2} ?

The integral of [sin(pi/3-x)]^{-2} is cot(pi/3-x)+C

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integral of [sin(pi/3-x)]^{-2}

\int [sin (\frac{π}{3} - x)]^{- 2} d x

cot (\frac{π}{3} - x) + C

Solution steps

\int [sin (\frac{π}{3} - x)]^{- 2} d x

Apply u-substitution

: \int - csc^{2} (u) d u

= \int - csc^{2} (u) d u

Take the constant out:

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

= - \int csc^{2} (u) d u

Use the common integral:

\int csc^{2} (u) d u = - cot (u)

= - (- cot (u))

Substitute back

u = \frac{π}{3} - x

= - (- cot (\frac{π}{3} - x))

Simplify

= cot (\frac{π}{3} - x)

Add a constant to the solution

= cot (\frac{π}{3} - x) + C

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What is the integral of [sin(pi/3-x)]^{-2} ?
The integral of [sin(pi/3-x)]^{-2} is cot(pi/3-x)+C