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Popular >

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

Solution

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

Solution

- \frac{1}{3} (ln \frac{tan ( \frac{x}{2} ) - 2}{3} + 1 - ln \frac{tan ( \frac{x}{2} ) - 2}{3} - 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{2}{1 + u ^{2} - 4 u} d u

= \int \frac{2}{1 + u ^{2} - 4 u} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{1 + u ^{2} - 4 u} d u

Complete the square

1 + u^{2} - 4 u : (u - 2)^{2} - 3

= 2 \cdot \int \frac{1}{( u - 2 ) ^{2} - 3} d u

Apply u-substitution

: \int \frac{1}{v ^{2} - 3} d v

= 2 \cdot \int \frac{1}{v ^{2} - 3} d v

Apply Integral Substitution

= 2 \cdot \int \frac{1}{3 ( w ^{2} - 1 )} d w

Take the constant out:

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

= 2 \cdot \frac{1}{3} \cdot \int \frac{1}{w ^{2} - 1} d w

Factor

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

= 2 \cdot \frac{1}{3} \cdot \int \frac{1}{- ( - w ^{2} + 1 )} d w

Take the constant out:

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

= 2 \cdot \frac{1}{3} (- \int \frac{1}{- w ^{2} + 1} d w)

Use the common integral:

\int \frac{1}{- w ^{2} + 1} d w = \frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}

= 2 \cdot \frac{1}{3} (- (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

Substitute back

= 2 \cdot \frac{1}{3} - \frac{ln \frac{t a n ( \frac{x}{2} ) - 2}{3} + 1}{2} - \frac{ln \frac{t a n ( \frac{x}{2} ) - 2}{3} - 1}{2}

Simplify

2 \cdot \frac{1}{3} - \frac{ln \frac{t a n ( \frac{x}{2} ) - 2}{3} + 1}{2} - \frac{ln \frac{t a n ( \frac{x}{2} ) - 2}{3} - 1}{2} : - \frac{1}{3} (ln \frac{tan ( \frac{x}{2} ) - 2}{3} + 1 - ln \frac{tan ( \frac{x}{2} ) - 2}{3} - 1)

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

Add a constant to the solution

= - \frac{1}{3} (ln \frac{tan ( \frac{x}{2} ) - 2}{3} + 1 - ln \frac{tan ( \frac{x}{2} ) - 2}{3} - 1) + C

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