What is the integral of-(4cos(t))/(3-10sin(t)) ?

The integral of-(4cos(t))/(3-10sin(t)) is 2/5 ln|3-10sin(t)|+C

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integral of-(4cos(t))/(3-10sin(t))

\int - \frac{4 cos ( t )}{3 - 10 sin ( t )} d t

\frac{2}{5} ln ∣ 3 - 10 sin (t) ∣ + C

Solution steps

\int - \frac{4 cos ( t )}{3 - 10 sin ( t )} d t

Take the constant out:

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

= - 4 \cdot \int \frac{cos ( t )}{3 - 10 sin ( t )} d t

Apply u-substitution

: \int - \frac{1}{10 u} d u

= - 4 \cdot \int - \frac{1}{10 u} d u

Take the constant out:

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

= - 4 (- \frac{1}{10} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 4 (- \frac{1}{10} ln ∣ u ∣)

Substitute back

u = 3 - 10 sin (t)

= - 4 (- \frac{1}{10} ln ∣ 3 - 10 sin (t) ∣)

Simplify

- 4 (- \frac{1}{10} ln ∣ 3 - 10 sin (t) ∣) : \frac{2}{5} ln ∣ 3 - 10 sin (t) ∣

= \frac{2}{5} ln ∣ 3 - 10 sin (t) ∣

Add a constant to the solution

= \frac{2}{5} ln ∣ 3 - 10 sin (t) ∣ + C

\int (2 x - 5)^{\frac{1}{2}} d x

f (x) = ln (x^{6} e^{x})

\int 24 + 2 x - x^{2} d x

What is the integral of-(4cos(t))/(3-10sin(t)) ?
The integral of-(4cos(t))/(3-10sin(t)) is 2/5 ln|3-10sin(t)|+C