What is the integral of (sin(2t))/(4-3cos(2t)) ?

The integral of (sin(2t))/(4-3cos(2t)) is 1/6 ln|6sin^2(t)+1|+C

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integral of (sin(2t))/(4-3cos(2t))

\int \frac{sin ( 2 t )}{4 - 3 cos ( 2 t )} d t

\frac{1}{6} ln 6 sin^{2} (t) + 1 + C

Solution steps

\int \frac{sin ( 2 t )}{4 - 3 cos ( 2 t )} d t

Rewrite using trig identities

= \int \frac{sin ( 2 t )}{6 sin ^{2} ( t ) + 1} d t

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} ln ∣ u ∣

Substitute back

u = 6 sin^{2} (t) + 1

= \frac{1}{6} ln 6 sin^{2} (t) + 1

Add a constant to the solution

= \frac{1}{6} ln 6 sin^{2} (t) + 1 + C

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What is the integral of (sin(2t))/(4-3cos(2t)) ?
The integral of (sin(2t))/(4-3cos(2t)) is 1/6 ln|6sin^2(t)+1|+C