What is the integral of 2sec^6(10x) ?

The integral of 2sec^6(10x) is 1/5 (tan(10x)+2/3 tan^3(10x)+1/5 tan^5(10x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 2sec^6(10x)

\int 2 sec^{6} (10 x) d x

\frac{1}{5} (tan (10 x) + \frac{2}{3} tan^{3} (10 x) + \frac{1}{5} tan^{5} (10 x)) + C

Solution steps

\int 2 sec^{6} (10 x) d x

Take the constant out:

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

= 2 \cdot \int sec^{6} (10 x) d x

Apply u-substitution

= 2 \cdot \int sec^{6} (u) \frac{1}{10} d u

Take the constant out:

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

= 2 \cdot \frac{1}{10} \cdot \int sec^{6} (u) d u

Rewrite using trig identities

= 2 \cdot \frac{1}{10} \cdot \int sec^{2} (u) (1 + tan^{2} (u))^{2} d u

Apply u-substitution

= 2 \cdot \frac{1}{10} \cdot \int (1 + v^{2})^{2} d v

Expand

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

= 2 \cdot \frac{1}{10} \cdot \int 1 + 2 v^{2} + v^{4} d v

Apply the Sum Rule:

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

= 2 \cdot \frac{1}{10} (\int 1 d v + \int 2 v^{2} d v + \int v^{4} d v)

\int 1 d v = v

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

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

= 2 \cdot \frac{1}{10} (v + \frac{2 v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= 2 \cdot \frac{1}{10} (tan (10 x) + \frac{2 tan ^{3} ( 10 x )}{3} + \frac{tan ^{5} ( 10 x )}{5})

Simplify

2 \cdot \frac{1}{10} (tan (10 x) + \frac{2 tan ^{3} ( 10 x )}{3} + \frac{tan ^{5} ( 10 x )}{5}) : \frac{1}{5} (tan (10 x) + \frac{2}{3} tan^{3} (10 x) + \frac{1}{5} tan^{5} (10 x))

= \frac{1}{5} (tan (10 x) + \frac{2}{3} tan^{3} (10 x) + \frac{1}{5} tan^{5} (10 x))

Add a constant to the solution

= \frac{1}{5} (tan (10 x) + \frac{2}{3} tan^{3} (10 x) + \frac{1}{5} tan^{5} (10 x)) + C

What is the integral of 2sec^6(10x) ?
The integral of 2sec^6(10x) is 1/5 (tan(10x)+2/3 tan^3(10x)+1/5 tan^5(10x))+C