What is the integral of (sin(2x))/(18+cos^2(x)) ?

The integral of (sin(2x))/(18+cos^2(x)) is -ln|cos(2x)+37|+C

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integral of (sin(2x))/(18+cos^2(x))

\int \frac{sin ( 2 x )}{18 + cos ^{2} ( x )} d x

- ln ∣ cos (2 x) + 37 ∣ + C

Solution steps

\int \frac{sin ( 2 x )}{18 + cos ^{2} ( x )} d x

Rewrite using trig identities

= \int \frac{2 sin ( 2 x )}{cos ( 2 x ) + 37} d x

Take the constant out:

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

= 2 \cdot \int \frac{sin ( 2 x )}{cos ( 2 x ) + 37} d x

Apply u-substitution

: \int - \frac{1}{2 u} d u

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

Take the constant out:

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

= 2 (- \frac{1}{2} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 (- \frac{1}{2} ln ∣ u ∣)

Substitute back

u = cos (2 x) + 37

= 2 (- \frac{1}{2} ln ∣ cos (2 x) + 37 ∣)

Simplify

2 (- \frac{1}{2} ln ∣ cos (2 x) + 37 ∣) : - ln ∣ cos (2 x) + 37 ∣

= - ln ∣ cos (2 x) + 37 ∣

Add a constant to the solution

= - ln ∣ cos (2 x) + 37 ∣ + C

\frac{d}{d x} (\frac{4 x ^{5} - 5 x ^{4} + 6 x - 2}{x ^{4}})

y^{^{'}} + 3 y + 5 = 0

\frac{d}{d x} (3 x^{2} + 12 x + 6)

Y = (x^{\frac{1}{2}}), Y = 6 - x, Y = 0

\int e^{x - 3} d x

What is the integral of (sin(2x))/(18+cos^2(x)) ?
The integral of (sin(2x))/(18+cos^2(x)) is -ln|cos(2x)+37|+C