What is the integral of 4/(15sin(x)-8cos(x)) ?

The integral of 4/(15sin(x)-8cos(x)) is -4/17 (ln| 8/17 tan(x/2)+32/17 |-ln| 8/17 tan(x/2)-2/17 |)+C

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integral of 4/(15sin(x)-8cos(x))

\int \frac{4}{15 sin ( x ) - 8 cos ( x )} d x

- \frac{4}{17} (ln \frac{8}{17} tan (\frac{x}{2}) + \frac{32}{17} - ln \frac{8}{17} tan (\frac{x}{2}) - \frac{2}{17}) + C

Solution steps

\int \frac{4}{15 sin ( x ) - 8 cos ( x )} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{15 sin ( x ) - 8 cos ( x )} d x

Apply u-substitution

: \int \frac{1}{15 u - 4 ( - u ^{2} + 1 )} d u

= 4 \cdot \int \frac{1}{15 u - 4 ( - u ^{2} + 1 )} d u

Complete the square

15 u - 4 (- u^{2} + 1) : 4 (u + \frac{15}{8})^{2} - \frac{289}{16}

= 4 \cdot \int \frac{1}{4 ( u + \frac{15}{8} ) ^{2} - \frac{289}{16}} d u

Apply u-substitution

: \int \frac{16}{64 v ^{2} - 289} d v

= 4 \cdot \int \frac{16}{64 v ^{2} - 289} d v

Take the constant out:

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

= 4 \cdot 16 \cdot \int \frac{1}{64 v ^{2} - 289} d v

Apply Integral Substitution

= 4 \cdot 16 \cdot \int \frac{1}{136 ( 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 16 \cdot \frac{1}{136} \cdot \int \frac{1}{w ^{2} - 1} d w

Factor

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

= 4 \cdot 16 \cdot \frac{1}{136} \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 16 \cdot \frac{1}{136} (- \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 16 \cdot \frac{1}{136} (- (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

Substitute back

= 4 \cdot 16 \cdot \frac{1}{136} (- (\frac{ln \frac{8}{17} ( tan ( \frac{x}{2} ) + \frac{15}{8} ) + 1}{2} - \frac{ln \frac{8}{17} ( tan ( \frac{x}{2} ) + \frac{15}{8} ) - 1}{2}))

Simplify

4 \cdot 16 \cdot \frac{1}{136} (- (\frac{ln \frac{8}{17} ( tan ( \frac{x}{2} ) + \frac{15}{8} ) + 1}{2} - \frac{ln \frac{8}{17} ( tan ( \frac{x}{2} ) + \frac{15}{8} ) - 1}{2})) : - \frac{4}{17} (ln \frac{8}{17} tan (\frac{x}{2}) + \frac{32}{17} - ln \frac{8}{17} tan (\frac{x}{2}) - \frac{2}{17})

= - \frac{4}{17} (ln \frac{8}{17} tan (\frac{x}{2}) + \frac{32}{17} - ln \frac{8}{17} tan (\frac{x}{2}) - \frac{2}{17})

Add a constant to the solution

= - \frac{4}{17} (ln \frac{8}{17} tan (\frac{x}{2}) + \frac{32}{17} - ln \frac{8}{17} tan (\frac{x}{2}) - \frac{2}{17}) + C

y = 4 - x^{2}

\frac{d y}{d x} = \frac{x ^{2} - 1}{y ^{2} + 1}, y (- 1) = 1

n \to \infty lim (\frac{n}{- n})

x \to 3 lim (\frac{3 x - 5}{3 - x})

\frac{d ^{2} x}{d t ^{2}} = - k \cdot x

What is the integral of 4/(15sin(x)-8cos(x)) ?
The integral of 4/(15sin(x)-8cos(x)) is -4/17 (ln| 8/17 tan(x/2)+32/17 |-ln| 8/17 tan(x/2)-2/17 |)+C