What is the integral of (2cos(x))/(-2+6sin(x)) ?

The integral of (2cos(x))/(-2+6sin(x)) is 1/3 ln|-2+6sin(x)|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2cos(x))/(-2+6sin(x))

\int \frac{2 cos ( x )}{- 2 + 6 sin ( x )} d x

\frac{1}{3} ln ∣ - 2 + 6 sin (x) ∣ + C

Solution steps

\int \frac{2 cos ( x )}{- 2 + 6 sin ( x )} d x

Take the constant out:

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

= 2 \cdot \int \frac{cos ( x )}{- 2 + 6 sin ( x )} d x

Apply u-substitution

: \int \frac{1}{6 u} d u

= 2 \cdot \int \frac{1}{6 u} d u

Take the constant out:

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

= 2 \cdot \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{6} ln ∣ u ∣

Substitute back

u = - 2 + 6 sin (x)

= 2 \cdot \frac{1}{6} ln ∣ - 2 + 6 sin (x) ∣

Simplify

2 \cdot \frac{1}{6} ln ∣ - 2 + 6 sin (x) ∣ : \frac{1}{3} ln ∣ - 2 + 6 sin (x) ∣

= \frac{1}{3} ln ∣ - 2 + 6 sin (x) ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ - 2 + 6 sin (x) ∣ + C

\frac{16 x}{x ^{2} + 16}

7 x^{2} e^{x}

f (x) = - 4 (2 x + 7)^{- 5} (2), at x = 2

\int \frac{1}{x 4 x + 9} d x

cos^{5} (x)

What is the integral of (2cos(x))/(-2+6sin(x)) ?
The integral of (2cos(x))/(-2+6sin(x)) is 1/3 ln|-2+6sin(x)|+C