What is the integral of 7tan^5(x)sec^4(x) ?

The integral of 7tan^5(x)sec^4(x) is 7((tan^6(x))/6+(tan^8(x))/8)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 7tan^5(x)sec^4(x)

\int 7 tan^{5} (x) sec^{4} (x) d x

7 (\frac{tan ^{6} ( x )}{6} + \frac{tan ^{8} ( x )}{8}) + C

Solution steps

\int 7 tan^{5} (x) sec^{4} (x) d x

Take the constant out:

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

= 7 \cdot \int tan^{5} (x) sec^{4} (x) d x

Rewrite using trig identities

= 7 \cdot \int (1 + tan^{2} (x)) sec^{2} (x) tan^{5} (x) d x

Apply u-substitution

: \int u^{5} (1 + u^{2}) d u

= 7 \cdot \int u^{5} (1 + u^{2}) d u

Expand

u^{5} (1 + u^{2}) : u^{5} + u^{7}

= 7 \cdot \int u^{5} + u^{7} 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

= 7 (\int u^{5} d u + \int u^{7} d u)

\int u^{5} d u = \frac{u ^{6}}{6}

\int u^{7} d u = \frac{u ^{8}}{8}

= 7 (\frac{u ^{6}}{6} + \frac{u ^{8}}{8})

Substitute back

u = tan (x)

= 7 (\frac{tan ^{6} ( x )}{6} + \frac{tan ^{8} ( x )}{8})

Add a constant to the solution

= 7 (\frac{tan ^{6} ( x )}{6} + \frac{tan ^{8} ( x )}{8}) + C

t \to 3 lim (1 - 2 t)

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

\int (2^{x} + 3^{x})^{2} d x

\frac{d y}{d t} = 6 y + t e^{6 t}

\frac{d}{d x} (x^{4} - 3 x^{2} + 5 x + 2)

What is the integral of 7tan^5(x)sec^4(x) ?
The integral of 7tan^5(x)sec^4(x) is 7((tan^6(x))/6+(tan^8(x))/8)+C