What is the integral from 0 to pi/2 of 1/(1-sin(x)) ?

The integral from 0 to pi/2 of 1/(1-sin(x)) is diverges

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integral from 0 to pi/2 of 1/(1-sin(x))

\int_{0}^{\frac{π}{2}} \frac{1}{1 - sin ( x )} d x

diverges

Solution steps

\int_{0}^{\frac{π}{2}} \frac{1}{1 - sin ( x )} d x

Apply u-substitution

: \int_{0}^{1} \frac{2}{1 + u ^{2} - 2 u} d u

= \int_{0}^{1} \frac{2}{1 + u ^{2} - 2 u} d u

Take the constant out:

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

= 2 \cdot \int_{0}^{1} \frac{1}{1 + u ^{2} - 2 u} d u

Complete the square

1 + u^{2} - 2 u : (u - 1)^{2}

= 2 \cdot \int_{0}^{1} \frac{1}{( u - 1 ) ^{2}} d u

Apply u-substitution

: \int_{- 1}^{0} \frac{1}{v ^{2}} d v

= 2 \cdot \int_{- 1}^{0} \frac{1}{v ^{2}} d v

Apply the Power Rule

: [- \frac{1}{v}]_{- 1}^{0}

= 2 [- \frac{1}{v}]_{- 1}^{0}

Compute the boundaries:

diverges

= diverges

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What is the integral from 0 to pi/2 of 1/(1-sin(x)) ?
The integral from 0 to pi/2 of 1/(1-sin(x)) is diverges