What is the integral of (sin(2x))/(33+cos^2(x)) ?

The integral of (sin(2x))/(33+cos^2(x)) is -ln|cos(2x)+67|+C

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integral of (sin(2x))/(33+cos^2(x))

\int \frac{sin ( 2 x )}{33 + cos ^{2} ( x )} d x

- ln ∣ cos (2 x) + 67 ∣ + C

Solution steps

\int \frac{sin ( 2 x )}{33 + cos ^{2} ( x )} d x

Rewrite using trig identities

= \int \frac{2 sin ( 2 x )}{cos ( 2 x ) + 67} 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 ) + 67} 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) + 67

= 2 (- \frac{1}{2} ln ∣ cos (2 x) + 67 ∣)

Simplify

2 (- \frac{1}{2} ln ∣ cos (2 x) + 67 ∣) : - ln ∣ cos (2 x) + 67 ∣

= - ln ∣ cos (2 x) + 67 ∣

Add a constant to the solution

= - ln ∣ cos (2 x) + 67 ∣ + C

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

x^{2} + 36

x \to 1 + lim (x^{\frac{1}{( x - 1 )}})

\int cos^{9} (x) tan^{3} (x) d x

25 y^{^{''}} - 40 y^{^{'}} + 16 y = 0, y (0) = 2, y^{^{'}} (0) = 3

What is the integral of (sin(2x))/(33+cos^2(x)) ?
The integral of (sin(2x))/(33+cos^2(x)) is -ln|cos(2x)+67|+C