What is the integral of (sin(2x))/(14+cos^2(x)) ?

The integral of (sin(2x))/(14+cos^2(x)) is -ln|cos(2x)+29|+C

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integral of (sin(2x))/(14+cos^2(x))

\int \frac{sin ( 2 x )}{14 + cos ^{2} ( x )} d x

- ln ∣ cos (2 x) + 29 ∣ + C

Solution steps

\int \frac{sin ( 2 x )}{14 + cos ^{2} ( x )} d x

Rewrite using trig identities

= \int \frac{2 sin ( 2 x )}{cos ( 2 x ) + 29} 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 ) + 29} 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) + 29

= 2 (- \frac{1}{2} ln ∣ cos (2 x) + 29 ∣)

Simplify

2 (- \frac{1}{2} ln ∣ cos (2 x) + 29 ∣) : - ln ∣ cos (2 x) + 29 ∣

= - ln ∣ cos (2 x) + 29 ∣

Add a constant to the solution

= - ln ∣ cos (2 x) + 29 ∣ + C

y^{^{'}} + \frac{2}{x} y = y^{2}, y (1) = 3

\frac{d}{d x} (\frac{a}{2 y})

t^{3} y^{^{'}} + 4 t^{2} y = 0

\int \frac{3 sin ^{3} ( x )}{cos ( x )} d x

\frac{\partial}{\partial x} (2 x^{2} + 3 y^{2} + z^{2})

What is the integral of (sin(2x))/(14+cos^2(x)) ?
The integral of (sin(2x))/(14+cos^2(x)) is -ln|cos(2x)+29|+C