What is the integral of (10sin(x))/(-3+6cos(x)) ?

The integral of (10sin(x))/(-3+6cos(x)) is -5/3 ln|-3+6cos(x)|+C

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integral of (10sin(x))/(-3+6cos(x))

\int \frac{10 sin ( x )}{- 3 + 6 cos ( x )} d x

- \frac{5}{3} ln ∣ - 3 + 6 cos (x) ∣ + C

Solution steps

\int \frac{10 sin ( x )}{- 3 + 6 cos ( x )} d x

Take the constant out:

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

= 10 \cdot \int \frac{sin ( x )}{- 3 + 6 cos ( x )} d x

Apply u-substitution

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

= 10 \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

= 10 (- \frac{1}{6} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 10 (- \frac{1}{6} ln ∣ u ∣)

Substitute back

u = - 3 + 6 cos (x)

= 10 (- \frac{1}{6} ln ∣ - 3 + 6 cos (x) ∣)

Simplify

10 (- \frac{1}{6} ln ∣ - 3 + 6 cos (x) ∣) : - \frac{5}{3} ln ∣ - 3 + 6 cos (x) ∣

= - \frac{5}{3} ln ∣ - 3 + 6 cos (x) ∣

Add a constant to the solution

= - \frac{5}{3} ln ∣ - 3 + 6 cos (x) ∣ + C

\frac{d}{d x} ((∣ x ∣) d x)

\frac{1}{3 x}

x^{\frac{6}{7}}

\int x^{8} e^{x^{9}} d x

\frac{d}{d x} (\frac{x ^{2}}{4 ^{x}})

What is the integral of (10sin(x))/(-3+6cos(x)) ?
The integral of (10sin(x))/(-3+6cos(x)) is -5/3 ln|-3+6cos(x)|+C