What is the integral of (sin(2x))/(43+cos^2(x)) ?

The integral of (sin(2x))/(43+cos^2(x)) is -ln|cos(2x)+87|+C

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integral of (sin(2x))/(43+cos^2(x))

\int \frac{sin ( 2 x )}{43 + cos ^{2} ( x )} d x

- ln ∣ cos (2 x) + 87 ∣ + C

Solution steps

\int \frac{sin ( 2 x )}{43 + cos ^{2} ( x )} d x

Rewrite using trig identities

= \int \frac{2 sin ( 2 x )}{cos ( 2 x ) + 87} 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 ) + 87} d x

Apply u-substitution

= 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) + 87

= 2 (- \frac{1}{2} ln ∣ cos (2 x) + 87 ∣)

Simplify

2 (- \frac{1}{2} ln ∣ cos (2 x) + 87 ∣) : - ln ∣ cos (2 x) + 87 ∣

= - ln ∣ cos (2 x) + 87 ∣

Add a constant to the solution

= - ln ∣ cos (2 x) + 87 ∣ + C

y^{^{'}} + 4 y = cos (x)

a re a 6 x - 18, [2, 6]

\int - 9 csc^{2} (x) d x

d er i v a t i v e g (x) = 9 - x^{2}

t an g e n t f (x) = - 4 - x^{2}

What is the integral of (sin(2x))/(43+cos^2(x)) ?
The integral of (sin(2x))/(43+cos^2(x)) is -ln|cos(2x)+87|+C