What is the integral of (sin(4x))/(cos(2x)) ?

The integral of (sin(4x))/(cos(2x)) is -cos(2x)+C

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integral of (sin(4x))/(cos(2x))

\int \frac{sin ( 4 x )}{cos ( 2 x )} d x

- cos (2 x) + C

Solution steps

\int \frac{sin ( 4 x )}{cos ( 2 x )} d x

Rewrite using trig identities

= \int sin (4 x) sec (2 x) d x

Apply u-substitution

: \int \frac{1}{2} sin (2 u) sec (u) d u

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

Take the constant out:

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

= \frac{1}{2} \cdot \int sin (2 u) sec (u) d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int sec (u) \cdot 2 cos (u) sin (u) d u

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int sec (u) cos (u) sin (u) d u

Express with sin, cos

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

Use the common integral:

\int sin (u) d u = - cos (u)

= \frac{1}{2} \cdot 2 (- cos (u))

Substitute back

u = 2 x

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

Simplify

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

= - cos (2 x)

Add a constant to the solution

= - cos (2 x) + C

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

x \to + 0 + lim (arctan (x))

x^{4} - 5 x^{3} + 2

\frac{\partial}{\partial x} (x^{2} y^{3})

(4, - 2), (1, 1)

What is the integral of (sin(4x))/(cos(2x)) ?
The integral of (sin(4x))/(cos(2x)) is -cos(2x)+C