What is the integral of 1/(5sin(x)+3) ?

The integral of 1/(5sin(x)+3) is -1/4 (ln| 9/4+3/4 tan(x/2)|-ln| 1/4+3/4 tan(x/2)|)+C

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integral of 1/(5sin(x)+3)

\int \frac{1}{5 sin ( x ) + 3} d x

- \frac{1}{4} (ln \frac{9}{4} + \frac{3}{4} tan (\frac{x}{2}) - ln \frac{1}{4} + \frac{3}{4} tan (\frac{x}{2})) + C

Solution steps

\int \frac{1}{5 sin ( x ) + 3} d x

Apply u-substitution

= \int \frac{2}{10 u + 3 ( 1 + u ^{2} )} 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}{10 u + 3 ( 1 + u ^{2} )} d u

Expand

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

= 2 \cdot \int \frac{1}{10 u + 3 + 3 u ^{2}} d u

Complete the square

10 u + 3 + 3 u^{2} : 3 (u + \frac{5}{3})^{2} - \frac{16}{3}

= 2 \cdot \int \frac{1}{3 ( u + \frac{5}{3} ) ^{2} - \frac{16}{3}} d u

Apply u-substitution

= 2 \cdot \int \frac{3}{9 v ^{2} - 16} d v

Take the constant out:

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

= 2 \cdot 3 \cdot \int \frac{1}{9 v ^{2} - 16} d v

Apply Integral Substitution

= 2 \cdot 3 \cdot \int \frac{1}{12 ( 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 3 \cdot \frac{1}{12} \cdot \int \frac{1}{w ^{2} - 1} d w

Factor

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

= 2 \cdot 3 \cdot \frac{1}{12} \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 3 \cdot \frac{1}{12} (- \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 3 \cdot \frac{1}{12} (- (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

Substitute back

= 2 \cdot 3 \cdot \frac{1}{12} (- (\frac{ln \frac{3}{4} ( tan ( \frac{x}{2} ) + \frac{5}{3} ) + 1}{2} - \frac{ln \frac{3}{4} ( tan ( \frac{x}{2} ) + \frac{5}{3} ) - 1}{2}))

Simplify

2 \cdot 3 \cdot \frac{1}{12} (- (\frac{ln \frac{3}{4} ( tan ( \frac{x}{2} ) + \frac{5}{3} ) + 1}{2} - \frac{ln \frac{3}{4} ( tan ( \frac{x}{2} ) + \frac{5}{3} ) - 1}{2})) : - \frac{1}{4} (ln \frac{9}{4} + \frac{3}{4} tan (\frac{x}{2}) - ln \frac{1}{4} + \frac{3}{4} tan (\frac{x}{2}))

= - \frac{1}{4} (ln \frac{9}{4} + \frac{3}{4} tan (\frac{x}{2}) - ln \frac{1}{4} + \frac{3}{4} tan (\frac{x}{2}))

Add a constant to the solution

= - \frac{1}{4} (ln \frac{9}{4} + \frac{3}{4} tan (\frac{x}{2}) - ln \frac{1}{4} + \frac{3}{4} tan (\frac{x}{2})) + C

What is the integral of 1/(5sin(x)+3) ?
The integral of 1/(5sin(x)+3) is -1/4 (ln| 9/4+3/4 tan(x/2)|-ln| 1/4+3/4 tan(x/2)|)+C