What is the integral of 9sec^6(x) ?

The integral of 9sec^6(x) is 9(tan(x)+(2tan^3(x))/3+(tan^5(x))/5)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 9sec^6(x)

\int 9 sec^{6} (x) d x

9 (tan (x) + \frac{2 tan ^{3} ( x )}{3} + \frac{tan ^{5} ( x )}{5}) + C

Solution steps

\int 9 sec^{6} (x) d x

Take the constant out:

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

= 9 \cdot \int sec^{6} (x) d x

Rewrite using trig identities

= 9 \cdot \int sec^{2} (x) (1 + tan^{2} (x))^{2} d x

Apply u-substitution

: \int (1 + u^{2})^{2} d u

= 9 \cdot \int (1 + u^{2})^{2} d u

Expand

(1 + u^{2})^{2} : 1 + 2 u^{2} + u^{4}

= 9 \cdot \int 1 + 2 u^{2} + u^{4} d u

Apply the Sum Rule:

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

= 9 (\int 1 d u + \int 2 u^{2} d u + \int u^{4} d u)

\int 1 d u = u

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

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

= 9 (u + \frac{2 u ^{3}}{3} + \frac{u ^{5}}{5})

Substitute back

u = tan (x)

= 9 (tan (x) + \frac{2 tan ^{3} ( x )}{3} + \frac{tan ^{5} ( x )}{5})

Add a constant to the solution

= 9 (tan (x) + \frac{2 tan ^{3} ( x )}{3} + \frac{tan ^{5} ( x )}{5}) + C

n = 0 \sum \infty \frac{3}{7}

\int \frac{1}{2} cos (\frac{1}{2}) t d t

\frac{d}{d y} (e^{2 y} - y cos (x y))

e^{2 x}, 0

\frac{d}{d x} (x^{\frac{2}{3}} (x^{2} - 8))

What is the integral of 9sec^6(x) ?
The integral of 9sec^6(x) is 9(tan(x)+(2tan^3(x))/3+(tan^5(x))/5)+C