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

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

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

Factor

3 cos (x) - 4 sin (x) : - (- 3 cos (x) + 4 sin (x))

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

Take the constant out:

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

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

Apply u-substitution

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

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

Complete the square

- 3 (1 - u^{2}) + 8 u : 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

\frac{d ^{2}}{d x ^{2}} (\frac{e ^{x}}{x})

x \to - 3 - lim (\frac{∣ x + 3 ∣}{x + 3})

x \to - 8 - lim (\frac{x + 3}{x + 8})

\int_{0}^{1} 15 cos (x^{2}) d x

\int \frac{9}{x ( x ^{4} + 1 )} d x

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