What is the integral of 128tan^4(x/2) ?

The integral of 128tan^4(x/2) is 256(1/3 tan^3(x/2)+1/2 x-tan(1/2 x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 128tan^4(x/2)

\int 128 tan^{4} (\frac{x}{2}) d x

256 (\frac{1}{3} tan^{3} (\frac{x}{2}) + \frac{1}{2} x - tan (\frac{1}{2} x)) + C

Solution steps

\int 128 tan^{4} (\frac{x}{2}) d x

Take the constant out:

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

= 128 \cdot \int tan^{4} (\frac{x}{2}) d x

\frac{x}{2} = x \frac{1}{2}

= 128 \cdot \int tan^{4} (x \frac{1}{2}) d x

Apply u-substitution

: \int tan^{4} (u) \cdot 2 d u

= 128 \cdot \int tan^{4} (u) \cdot 2 d u

Take the constant out:

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

= 128 \cdot 2 \cdot \int tan^{4} (u) d u

Apply Integral Reduction

= 128 \cdot 2 (\frac{tan ^{3} ( u )}{3} - \int tan^{2} (u) d u)

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

= 128 \cdot 2 (\frac{tan ^{3} ( u )}{3} - (- u + tan (u)))

Substitute back

u = \frac{1}{2} x

= 128 \cdot 2 (\frac{tan ^{3} ( \frac{1}{2} x )}{3} - (- \frac{1}{2} x + tan (\frac{1}{2} x)))

Simplify

128 \cdot 2 (\frac{tan ^{3} ( \frac{1}{2} x )}{3} - (- \frac{1}{2} x + tan (\frac{1}{2} x))) : 256 (\frac{1}{3} tan^{3} (\frac{x}{2}) + \frac{1}{2} x - tan (\frac{1}{2} x))

= 256 (\frac{1}{3} tan^{3} (\frac{x}{2}) + \frac{1}{2} x - tan (\frac{1}{2} x))

Add a constant to the solution

= 256 (\frac{1}{3} tan^{3} (\frac{x}{2}) + \frac{1}{2} x - tan (\frac{1}{2} x)) + C

2 x^{2} y y^{^{'}} + y^{2} = 2

\frac{d}{d x} (- \frac{10}{( 1 + 5 x ) ^{2}})

\frac{\partial}{\partial x} (\frac{x + 4}{y + 6})

(4 x - 3)^{\frac{1}{2}}

x y^{^{'}} + y = 3 cos (x)

What is the integral of 128tan^4(x/2) ?
The integral of 128tan^4(x/2) is 256(1/3 tan^3(x/2)+1/2 x-tan(1/2 x))+C