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

The integral of (sin(2x))/(2cos^2(x)) is -1/2 ln(cos^2(x))+C

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

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

- \frac{1}{2} ln (cos^{2} (x)) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{sin ( 2 x )}{cos ^{2} ( x )} d x

Apply u-substitution

: \int - \frac{1}{u} d u

= \frac{1}{2} \cdot \int - \frac{1}{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} (- \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} (- ln ∣ u ∣)

Substitute back

u = cos^{2} (x)

= \frac{1}{2} (- ln cos^{2} (x))

Simplify

\frac{1}{2} (- ln cos^{2} (x)) : - \frac{1}{2} ln (cos^{2} (x))

= - \frac{1}{2} ln (cos^{2} (x))

Add a constant to the solution

= - \frac{1}{2} ln (cos^{2} (x)) + C

\frac{d}{d x} (- 32 sin (2 x))

1 + x - x^{2} - x^{3}, 1 - x, [- 2, 1]

\frac{\partial}{\partial y} (cos (yz))

(x + 2) sin (y) + (x cos (y)) y^{^{'}} = 0

x \to 0 + lim (ln (0) - \frac{1}{x})

What is the integral of (sin(2x))/(2cos^2(x)) ?
The integral of (sin(2x))/(2cos^2(x)) is -1/2 ln(cos^2(x))+C