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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

ln (6 - 7 x)

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

x \to - 5 lim (3 x^{3} + 15 x^{2} - 25)

n = 1 \sum \infty n^{2} e^{- n^{3}}

\frac{6}{s ^{2} + 9}

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