What is the integral of (sin(2x))/(26+cos^2(x)) ?

The integral of (sin(2x))/(26+cos^2(x)) is -ln|cos(2x)+53|+C

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integral of (sin(2x))/(26+cos^2(x))

\int \frac{sin ( 2 x )}{26 + cos ^{2} ( x )} d x

- ln ∣ cos (2 x) + 53 ∣ + C

Solution steps

\int \frac{sin ( 2 x )}{26 + cos ^{2} ( x )} d x

Rewrite using trig identities

= \int \frac{2 sin ( 2 x )}{cos ( 2 x ) + 53} 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 ) + 53} 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) + 53

= 2 (- \frac{1}{2} ln ∣ cos (2 x) + 53 ∣)

Simplify

2 (- \frac{1}{2} ln ∣ cos (2 x) + 53 ∣) : - ln ∣ cos (2 x) + 53 ∣

= - ln ∣ cos (2 x) + 53 ∣

Add a constant to the solution

= - ln ∣ cos (2 x) + 53 ∣ + C

\int \frac{3 x ^{2} + x + 4}{x ^{3} + x} d x

n = 1 \sum \infty \frac{1}{n ^{2} + 3 n}

f (x) = 270

y = x^{3} - 4 x^{2} + 3 x, y = x^{2} - x

\frac{d y}{d x} = x + 6 y

What is the integral of (sin(2x))/(26+cos^2(x)) ?
The integral of (sin(2x))/(26+cos^2(x)) is -ln|cos(2x)+53|+C