What is the integral of sec^8(x)tan(x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of sec^8(x)tan(x)

\int sec^{8} (x) tan (x) d x

\frac{1}{8} sec^{8} (x) + C

Solution steps

\int sec^{8} (x) tan (x) d x

Rewrite using trig identities

= \int \frac{sin ( x )}{cos ^{9} ( x )} d x

Apply u-substitution

= \int - \frac{1}{u ^{9}} d u

Take the constant out:

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

= - \int \frac{1}{u ^{9}} d u

Apply the Power Rule

= - (- \frac{1}{8 u ^{8}})

Substitute back

u = cos (x)

= - (- \frac{1}{8 cos ^{8} ( x )})

Simplify

- (- \frac{1}{8 cos ^{8} ( x )}) : \frac{1}{8} sec^{8} (x)

= \frac{1}{8} sec^{8} (x)

Add a constant to the solution

= \frac{1}{8} sec^{8} (x) + C

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