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

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

Solution steps

\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

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

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

Add a constant to the solution

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

x \to - \infty lim (0 x)

\int 2 x + 22 d x

f (x) = ln (x) cos (x)

\frac{d y}{d x} + 7 \frac{y}{( - 2 x + 200 )} = 30

\frac{d}{d x} (e^{x} cos (y) + ln (x^{2} y))

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