What is the integral of 1/(4sin(x)-3cos(x)) ?

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

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integral of 1/(4sin(x)-3cos(x))

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

- \frac{1}{5} (ln \frac{9}{5} + \frac{3}{5} tan (\frac{x}{2}) - ln - \frac{1}{5} + \frac{3}{5} tan (\frac{x}{2})) + C

Solution steps

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

Apply u-substitution

: \int \frac{2}{8 u - 3 ( 1 - u ^{2} )} d u

= \int \frac{2}{8 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}{8 u - 3 ( 1 - u ^{2} )} d u

Complete the square

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

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

Apply u-substitution

: \int \frac{3}{9 v ^{2} - 25} d v

= 2 \cdot \int \frac{3}{9 v ^{2} - 25} 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} - 25} d v

Apply Integral Substitution

= 2 \cdot 3 \cdot \int \frac{1}{15 ( 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}{15} \cdot \int \frac{1}{w ^{2} - 1} d w

Factor

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

= 2 \cdot 3 \cdot \frac{1}{15} \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}{15} (- \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}{15} (- (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

Substitute back

= 2 \cdot 3 \cdot \frac{1}{15} (- (\frac{ln \frac{3}{5} ( tan ( \frac{x}{2} ) + \frac{4}{3} ) + 1}{2} - \frac{ln \frac{3}{5} ( tan ( \frac{x}{2} ) + \frac{4}{3} ) - 1}{2}))

Simplify

2 \cdot 3 \cdot \frac{1}{15} (- (\frac{ln \frac{3}{5} ( tan ( \frac{x}{2} ) + \frac{4}{3} ) + 1}{2} - \frac{ln \frac{3}{5} ( tan ( \frac{x}{2} ) + \frac{4}{3} ) - 1}{2})) : - \frac{1}{5} (ln \frac{9}{5} + \frac{3}{5} tan (\frac{x}{2}) - ln - \frac{1}{5} + \frac{3}{5} tan (\frac{x}{2}))

= - \frac{1}{5} (ln \frac{9}{5} + \frac{3}{5} tan (\frac{x}{2}) - ln - \frac{1}{5} + \frac{3}{5} tan (\frac{x}{2}))

Add a constant to the solution

= - \frac{1}{5} (ln \frac{9}{5} + \frac{3}{5} tan (\frac{x}{2}) - ln - \frac{1}{5} + \frac{3}{5} tan (\frac{x}{2})) + C

\int_{1}^{\infty} \frac{1}{x ^{0}} d x

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

\int_{- 1}^{1} (e^{x})^{2} d x

\int (\frac{5}{t ^{2}} + 4 t^{2}) d t

\int \frac{1}{5 - x ^{2}} d x

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