What is the integral of tan^6(4x) ?

The integral of tan^6(4x) is 1/4 ((tan^5(4x))/5-(tan^3(4x))/3+tan(4x)-arctan(tan(4x)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of tan^6(4x)

\int tan^{6} (4 x) d x

\frac{1}{4} (\frac{tan ^{5} ( 4 x )}{5} - \frac{tan ^{3} ( 4 x )}{3} + tan (4 x) - arctan (tan (4 x))) + C

Solution steps

\int tan^{6} (4 x) d x

Apply u-substitution

: \int tan^{6} (u) \frac{1}{4} d u

= \int tan^{6} (u) \frac{1}{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 tan^{6} (u) d u

Apply u-substitution

: \int \frac{v ^{6}}{1 + v ^{2}} d v

= \frac{1}{4} \cdot \int \frac{v ^{6}}{1 + v ^{2}} d v

Long division

\frac{v ^{6}}{1 + v ^{2}} : v^{4} - v^{2} + 1 - \frac{1}{v ^{2} + 1}

= \frac{1}{4} \cdot \int v^{4} - v^{2} + 1 - \frac{1}{v ^{2} + 1} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (\int v^{4} d v - \int v^{2} d v + \int 1 d v - \int \frac{1}{v ^{2} + 1} d v)

\int v^{4} d v = \frac{v ^{5}}{5}

\int v^{2} d v = \frac{v ^{3}}{3}

\int 1 d v = v

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{4} (\frac{v ^{5}}{5} - \frac{v ^{3}}{3} + v - arctan (v))

Substitute back

= \frac{1}{4} (\frac{tan ^{5} ( 4 x )}{5} - \frac{tan ^{3} ( 4 x )}{3} + tan (4 x) - arctan (tan (4 x)))

Add a constant to the solution

= \frac{1}{4} (\frac{tan ^{5} ( 4 x )}{5} - \frac{tan ^{3} ( 4 x )}{3} + tan (4 x) - arctan (tan (4 x))) + C

\int \frac{5 y + 4}{5 y - 1} d y

\int \frac{1}{( x - 1 ) ( x - 2 )} d x

\int \frac{6 x ^{2} - 5}{x - 1} d x

\int 3 x^{2} (x^{3} - 4)^{3} d x

What is the integral of tan^6(4x) ?
The integral of tan^6(4x) is 1/4 ((tan^5(4x))/5-(tan^3(4x))/3+tan(4x)-arctan(tan(4x)))+C