What is the integral of (sin(2x))/(cos^3(2x)) ?

The integral of (sin(2x))/(cos^3(2x)) is 1/4 tan^2(2x)+C

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integral of (sin(2x))/(cos^3(2x))

\int \frac{sin ( 2 x )}{cos ^{3} ( 2 x )} d x

\frac{1}{4} tan^{2} (2 x) + C

Solution steps

\int \frac{sin ( 2 x )}{cos ^{3} ( 2 x )} d x

Apply u-substitution

: \int \frac{sin ( u )}{2 cos ^{3} ( u )} d u

= \int \frac{sin ( u )}{2 cos ^{3} ( 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 \frac{sin ( u )}{cos ^{3} ( u )} d u

Rewrite using trig identities

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

Apply u-substitution

: \int v d v

= \frac{1}{2} \cdot \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= \frac{1}{2} \cdot \frac{v ^{2}}{2}

Substitute back

= \frac{1}{2} \cdot \frac{tan ^{2} ( 2 x )}{2}

Simplify

\frac{1}{2} \cdot \frac{tan ^{2} ( 2 x )}{2} : \frac{1}{4} tan^{2} (2 x)

= \frac{1}{4} tan^{2} (2 x)

Add a constant to the solution

= \frac{1}{4} tan^{2} (2 x) + C

x, \frac{1}{x ^{2}}, 1, 3

\frac{d y}{d x} = - 5 y x, y (0) = 50

x 2^{x}

\int sin^{4} (x) \cdot cos^{5} (x) d x

\frac{d}{d x} (x (1 + x)^{- 1})

What is the integral of (sin(2x))/(cos^3(2x)) ?
The integral of (sin(2x))/(cos^3(2x)) is 1/4 tan^2(2x)+C