What is the integral of tan^7(x/2)sec^2(x/2) ?

The integral of tan^7(x/2)sec^2(x/2) is 1/4 tan^8(x/2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of tan^7(x/2)sec^2(x/2)

\int tan^{7} (\frac{x}{2}) sec^{2} (\frac{x}{2}) d x

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

Solution steps

\int tan^{7} (\frac{x}{2}) sec^{2} (\frac{x}{2}) d x

Apply u-substitution

: \int 2 u^{7} d u

= \int 2 u^{7} d u

Take the constant out:

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

= 2 \cdot \int u^{7} d u

Apply the Power Rule

: \frac{u ^{8}}{8}

= 2 \cdot \frac{u ^{8}}{8}

Substitute back

u = tan (\frac{x}{2})

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

Simplify

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

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

Add a constant to the solution

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

\int 1.8 t d t

\int e^{x} 49 - e^{2 x} d x

f (x) = ln (x - 4)

\int \frac{x ^{3}}{x ^{2} - 64} d x

f (x) = \frac{1}{x ^{2} + 1}, at x = 2

What is the integral of tan^7(x/2)sec^2(x/2) ?
The integral of tan^7(x/2)sec^2(x/2) is 1/4 tan^8(x/2)+C