What is the integral of (sin(2x))/(32+cos^2(x)) ?

The integral of (sin(2x))/(32+cos^2(x)) is -ln|cos(2x)+65|+C

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integral of (sin(2x))/(32+cos^2(x))

\int \frac{sin ( 2 x )}{32 + cos ^{2} ( x )} d x

- ln ∣ cos (2 x) + 65 ∣ + C

Solution steps

\int \frac{sin ( 2 x )}{32 + cos ^{2} ( x )} d x

Rewrite using trig identities

= \int \frac{2 sin ( 2 x )}{cos ( 2 x ) + 65} 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 ) + 65} 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) + 65

= 2 (- \frac{1}{2} ln ∣ cos (2 x) + 65 ∣)

Simplify

2 (- \frac{1}{2} ln ∣ cos (2 x) + 65 ∣) : - ln ∣ cos (2 x) + 65 ∣

= - ln ∣ cos (2 x) + 65 ∣

Add a constant to the solution

= - ln ∣ cos (2 x) + 65 ∣ + C

\frac{d P}{d t} = 0.16 P (1 - \frac{P}{1000}) - 30, P (0) = 200

6 cos (t) - 2 sin (t)

f (t) = t^{3} - 27 t + 5

\int_{0}^{1} \frac{5}{4 - x ^{2}} d x

y = ln (x^{4} - 4095), (8, 0)

What is the integral of (sin(2x))/(32+cos^2(x)) ?
The integral of (sin(2x))/(32+cos^2(x)) is -ln|cos(2x)+65|+C