What is the integral of (sin(2x))/(47+cos^2(x)) ?

The integral of (sin(2x))/(47+cos^2(x)) is -ln|cos(2x)+95|+C

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integral of (sin(2x))/(47+cos^2(x))

\int \frac{sin ( 2 x )}{47 + cos ^{2} ( x )} d x

- ln ∣ cos (2 x) + 95 ∣ + C

Solution steps

\int \frac{sin ( 2 x )}{47 + cos ^{2} ( x )} d x

Rewrite using trig identities

= \int \frac{2 sin ( 2 x )}{cos ( 2 x ) + 95} 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 ) + 95} 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) + 95

= 2 (- \frac{1}{2} ln ∣ cos (2 x) + 95 ∣)

Simplify

2 (- \frac{1}{2} ln ∣ cos (2 x) + 95 ∣) : - ln ∣ cos (2 x) + 95 ∣

= - ln ∣ cos (2 x) + 95 ∣

Add a constant to the solution

= - ln ∣ cos (2 x) + 95 ∣ + C

y = \frac{6 x}{x ^{2} + 1}, (- 1, - 3)

(3 x)^{^{'}}

\frac{\partial}{\partial x} (y^{2} cos (y))

x \to - 2 lim (x^{3} + 4 x^{2} - 5 x + 1)

What is the integral of (sin(2x))/(47+cos^2(x)) ?
The integral of (sin(2x))/(47+cos^2(x)) is -ln|cos(2x)+95|+C