What is the integral of 16tan^4(x)sec^6(x) ?

The integral of 16tan^4(x)sec^6(x) is 16((tan^5(x))/5+(2tan^7(x))/7+(tan^9(x))/9)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 16tan^4(x)sec^6(x)

\int 16 tan^{4} (x) sec^{6} (x) d x

16 (\frac{tan ^{5} ( x )}{5} + \frac{2 tan ^{7} ( x )}{7} + \frac{tan ^{9} ( x )}{9}) + C

Solution steps

\int 16 tan^{4} (x) sec^{6} (x) d x

Take the constant out:

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

= 16 \cdot \int tan^{4} (x) sec^{6} (x) d x

Rewrite using trig identities

= 16 \cdot \int sec^{2} (x) (1 + tan^{2} (x))^{2} tan^{4} (x) d x

Apply u-substitution

= 16 \cdot \int u^{4} (1 + u^{2})^{2} d u

Expand

u^{4} (1 + u^{2})^{2} : u^{4} + 2 u^{6} + u^{8}

= 16 \cdot \int u^{4} + 2 u^{6} + u^{8} 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

= 16 (\int u^{4} d u + \int 2 u^{6} d u + \int u^{8} d u)

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

\int 2 u^{6} d u = \frac{2 u ^{7}}{7}

\int u^{8} d u = \frac{u ^{9}}{9}

= 16 (\frac{u ^{5}}{5} + \frac{2 u ^{7}}{7} + \frac{u ^{9}}{9})

Substitute back

u = tan (x)

= 16 (\frac{tan ^{5} ( x )}{5} + \frac{2 tan ^{7} ( x )}{7} + \frac{tan ^{9} ( x )}{9})

Add a constant to the solution

= 16 (\frac{tan ^{5} ( x )}{5} + \frac{2 tan ^{7} ( x )}{7} + \frac{tan ^{9} ( x )}{9}) + C

t an g e n t f (x) = 38 - x^{2}, (- 6, 2)

t an g e n t y = 3 - x^{3} (0.3)

\int x^{2} (x^{3} - 1)^{2} d x

z \to 2 i lim (i z^{4} + 3 z^{2} - 10 i)

d er i v a t i v e f (x) = x^{2} - 8 x

What is the integral of 16tan^4(x)sec^6(x) ?
The integral of 16tan^4(x)sec^6(x) is 16((tan^5(x))/5+(2tan^7(x))/7+(tan^9(x))/9)+C