What is the integral of (sin(4x))/(cos^3(4x)) ?

The integral of (sin(4x))/(cos^3(4x)) is 1/8 tan^2(4x)+C

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integral of (sin(4x))/(cos^3(4x))

\int \frac{sin ( 4 x )}{cos ^{3} ( 4 x )} d x

\frac{1}{8} tan^{2} (4 x) + C

Solution steps

\int \frac{sin ( 4 x )}{cos ^{3} ( 4 x )} d x

Rewrite using trig identities

= \int \frac{tan ( 4 x )}{cos ^{2} ( 4 x )} d x

Use the basic trigonometric identity:

\frac{1}{cos ( x )} = sec (x)

= \int sec^{2} (4 x) tan (4 x) d x

Apply u-substitution

: \int \frac{u}{4} d u

= \int \frac{u}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{1}{4} \cdot \frac{u ^{2}}{2}

Substitute back

u = tan (4 x)

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

Simplify

\frac{1}{4} \cdot \frac{tan ^{2} ( 4 x )}{2} : \frac{1}{8} tan^{2} (4 x)

= \frac{1}{8} tan^{2} (4 x)

Add a constant to the solution

= \frac{1}{8} tan^{2} (4 x) + C

\frac{\partial}{\partial x} (k)

\frac{d}{d x} ((3 x - 3 x^{2}) tan (x))

\int \frac{1}{a x ^{2} + b x + c} d x

y^{^{'}} - (\frac{2 x}{1 + x ^{2}}) y = 0, y (0) = 4

\frac{d}{d x} (\frac{x ^{3}}{3} - x^{2} - 3 x)

What is the integral of (sin(4x))/(cos^3(4x)) ?
The integral of (sin(4x))/(cos^3(4x)) is 1/8 tan^2(4x)+C