What is the integral from 0 to pi/3 of (4sin(θ))/(1-4cos(θ)) ?

The integral from 0 to pi/3 of (4sin(θ))/(1-4cos(θ)) is -ln(3)

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integral from 0 to pi/3 of (4sin(θ))/(1-4cos(θ))

\int_{0}^{\frac{π}{3}} \frac{4 sin ( θ )}{1 - 4 cos ( θ )} d θ

- ln (3)

Decimal

- 1.09861 \dots

Solution steps

\int_{0}^{\frac{π}{3}} \frac{4 sin ( θ )}{1 - 4 cos ( θ )} d θ

Take the constant out:

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

= 4 \cdot \int_{0}^{\frac{π}{3}} \frac{sin ( θ )}{1 - 4 cos ( θ )} d θ

Apply u-substitution

= 4 \cdot \int_{- 3}^{- 1} \frac{1}{4 u} d u

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot \int_{- 3}^{- 1} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{4} [ln ∣ u ∣]_{- 3}^{- 1}

Simplify

4 \cdot \frac{1}{4} [ln ∣ u ∣]_{- 3}^{- 1} : [ln ∣ u ∣]_{- 3}^{- 1}

= [ln ∣ u ∣]_{- 3}^{- 1}

Compute the boundaries:

- ln (3)

= - ln (3)

\int_{1}^{2} x + 1 d x

\int_{- 7}^{7} x^{3} d x

\int_{1}^{3} r^{4} ln (r) d r

\int_{1}^{1} \frac{1}{t} d t

\int_{0}^{1} 3 e^{x} d x

What is the integral from 0 to pi/3 of (4sin(θ))/(1-4cos(θ)) ?
The integral from 0 to pi/3 of (4sin(θ))/(1-4cos(θ)) is -ln(3)