What is the integral of tan^4(7x) ?

The integral of tan^4(7x) is 1/21 tan^3(7x)+x-1/7 tan(7x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of tan^4(7x)

\int tan^{4} (7 x) d x

\frac{1}{21} tan^{3} (7 x) + x - \frac{1}{7} tan (7 x) + C

Solution steps

\int tan^{4} (7 x) d x

Apply u-substitution

: \int tan^{4} (u) \frac{1}{7} d u

= \int tan^{4} (u) \frac{1}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int tan^{4} (u) d u

Apply Integral Reduction

= \frac{1}{7} (\frac{tan ^{3} ( u )}{3} - \int tan^{2} (u) d u)

\int tan^{2} (u) d u = - u + tan (u)

= \frac{1}{7} (\frac{tan ^{3} ( u )}{3} - (- u + tan (u)))

Substitute back

u = 7 x

= \frac{1}{7} (\frac{tan ^{3} ( 7 x )}{3} - (- 7 x + tan (7 x)))

Simplify

\frac{1}{7} (\frac{tan ^{3} ( 7 x )}{3} - (- 7 x + tan (7 x))) : \frac{1}{21} tan^{3} (7 x) + x - \frac{1}{7} tan (7 x)

= \frac{1}{21} tan^{3} (7 x) + x - \frac{1}{7} tan (7 x)

Add a constant to the solution

= \frac{1}{21} tan^{3} (7 x) + x - \frac{1}{7} tan (7 x) + C

\frac{\partial}{\partial x} (cos (x) sin (x) - x y)

\frac{\partial}{\partial x} (4)

a x + b

\frac{( 3 x )}{( 1 + x ^{2} )}

\int tan (5 v) sec (5 v) d v

What is the integral of tan^4(7x) ?
The integral of tan^4(7x) is 1/21 tan^3(7x)+x-1/7 tan(7x)+C