What is the integral of 6/(3sin(x)-4cos(x)) ?

The integral of 6/(3sin(x)-4cos(x)) is -6/5 (ln| 4/5 tan(x/2)+8/5 |-ln| 4/5 tan(x/2)-2/5 |)+C

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integral of 6/(3sin(x)-4cos(x))

\int \frac{6}{3 sin ( x ) - 4 cos ( x )} d x

- \frac{6}{5} (ln \frac{4}{5} tan (\frac{x}{2}) + \frac{8}{5} - ln \frac{4}{5} tan (\frac{x}{2}) - \frac{2}{5}) + C

Solution steps

\int \frac{6}{3 sin ( x ) - 4 cos ( x )} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{3 sin ( x ) - 4 cos ( x )} d x

Apply u-substitution

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

= 6 \cdot \int \frac{1}{3 u - 2 ( - u ^{2} + 1 )} d u

Complete the square

3 u - 2 (- u^{2} + 1) : 2 (u + \frac{3}{4})^{2} - \frac{25}{8}

= 6 \cdot \int \frac{1}{2 ( u + \frac{3}{4} ) ^{2} - \frac{25}{8}} d u

Apply u-substitution

: \int \frac{8}{16 v ^{2} - 25} d v

= 6 \cdot \int \frac{8}{16 v ^{2} - 25} d v

Take the constant out:

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

= 6 \cdot 8 \cdot \int \frac{1}{16 v ^{2} - 25} d v

Apply Integral Substitution

= 6 \cdot 8 \cdot \int \frac{1}{20 ( w ^{2} - 1 )} d w

Take the constant out:

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

= 6 \cdot 8 \cdot \frac{1}{20} \cdot \int \frac{1}{w ^{2} - 1} d w

Factor

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

= 6 \cdot 8 \cdot \frac{1}{20} \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

= 6 \cdot 8 \cdot \frac{1}{20} (- \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}

= 6 \cdot 8 \cdot \frac{1}{20} (- (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

Substitute back

= 6 \cdot 8 \cdot \frac{1}{20} (- (\frac{ln \frac{4}{5} ( tan ( \frac{x}{2} ) + \frac{3}{4} ) + 1}{2} - \frac{ln \frac{4}{5} ( tan ( \frac{x}{2} ) + \frac{3}{4} ) - 1}{2}))

Simplify

6 \cdot 8 \cdot \frac{1}{20} (- (\frac{ln \frac{4}{5} ( tan ( \frac{x}{2} ) + \frac{3}{4} ) + 1}{2} - \frac{ln \frac{4}{5} ( tan ( \frac{x}{2} ) + \frac{3}{4} ) - 1}{2})) : - \frac{6}{5} (ln \frac{4}{5} tan (\frac{x}{2}) + \frac{8}{5} - ln \frac{4}{5} tan (\frac{x}{2}) - \frac{2}{5})

= - \frac{6}{5} (ln \frac{4}{5} tan (\frac{x}{2}) + \frac{8}{5} - ln \frac{4}{5} tan (\frac{x}{2}) - \frac{2}{5})

Add a constant to the solution

= - \frac{6}{5} (ln \frac{4}{5} tan (\frac{x}{2}) + \frac{8}{5} - ln \frac{4}{5} tan (\frac{x}{2}) - \frac{2}{5}) + C

\int \frac{2}{x ^{2}} - \frac{x ^{2}}{2} d x

n = 1 \sum \infty n e^{- 7 n}

\frac{d}{d x} (arctan^{t a n (x)} (x))

\int \frac{x}{1 + x ^{3}} d x

\frac{d}{d x} (- 8 x - cos (x))

What is the integral of 6/(3sin(x)-4cos(x)) ?
The integral of 6/(3sin(x)-4cos(x)) is -6/5 (ln| 4/5 tan(x/2)+8/5 |-ln| 4/5 tan(x/2)-2/5 |)+C