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Popular >

integral of 1/(2-3sin(x))

Solution

\int \frac{1}{2 - 3 sin ( x )} d x

Solution

- \frac{1}{5} (ln \frac{2 tan ( \frac{x}{2} ) - 3}{5} + 1 - ln \frac{2 tan ( \frac{x}{2} ) - 3}{5} - 1) + C

Solution steps

\int \frac{1}{2 - 3 sin ( x )} d x

Apply u-substitution

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

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

Complete the square

u^{2} + 1 - 3 u : (u - \frac{3}{2})^{2} - \frac{5}{4}

= \int \frac{1}{( u - \frac{3}{2} ) ^{2} - \frac{5}{4}} d u

Apply u-substitution

: \int \frac{4}{4 v ^{2} - 5} d v

= \int \frac{4}{4 v ^{2} - 5} d v

Take the constant out:

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

= 4 \cdot \int \frac{1}{4 v ^{2} - 5} d v

Apply Integral Substitution

= 4 \cdot \int \frac{1}{2 5 ( w ^{2} - 1 )} d w

Take the constant out:

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

= 4 \cdot \frac{1}{2 5} \cdot \int \frac{1}{w ^{2} - 1} d w

Factor

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

= 4 \cdot \frac{1}{2 5} \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

= 4 \cdot \frac{1}{2 5} (- \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}

= 4 \cdot \frac{1}{2 5} (- (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

Substitute back

= 4 \cdot \frac{1}{2 5} - \frac{ln \frac{2}{5} ( tan ( \frac{x}{2} ) - \frac{3}{2} ) + 1}{2} - \frac{ln \frac{2}{5} ( tan ( \frac{x}{2} ) - \frac{3}{2} ) - 1}{2}

Simplify

4 \cdot \frac{1}{2 5} - \frac{ln \frac{2}{5} ( tan ( \frac{x}{2} ) - \frac{3}{2} ) + 1}{2} - \frac{ln \frac{2}{5} ( tan ( \frac{x}{2} ) - \frac{3}{2} ) - 1}{2} : - \frac{1}{5} (ln \frac{2 tan ( \frac{x}{2} ) - 3}{5} + 1 - ln \frac{2 tan ( \frac{x}{2} ) - 3}{5} - 1)

= - \frac{1}{5} (ln \frac{2 tan ( \frac{x}{2} ) - 3}{5} + 1 - ln \frac{2 tan ( \frac{x}{2} ) - 3}{5} - 1)

Add a constant to the solution

= - \frac{1}{5} (ln \frac{2 tan ( \frac{x}{2} ) - 3}{5} + 1 - ln \frac{2 tan ( \frac{x}{2} ) - 3}{5} - 1) + C

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