What is the integral of 41tan^5(x)sec^4(x) ?

The integral of 41tan^5(x)sec^4(x) is 41((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 41tan^5(x)sec^4(x)

\int 41 tan^{5} (x) sec^{4} (x) d x

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

Solution steps

\int 41 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

= 41 \cdot \int tan^{5} (x) sec^{4} (x) d x

Rewrite using trig identities

= 41 \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

= 41 \cdot \int u^{5} (1 + u^{2}) d u

Expand

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

= 41 \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

= 41 (\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}

= 41 (\frac{u ^{6}}{6} + \frac{u ^{8}}{8})

Substitute back

u = tan (x)

= 41 (\frac{tan ^{6} ( x )}{6} + \frac{tan ^{8} ( x )}{8})

Add a constant to the solution

= 41 (\frac{tan ^{6} ( x )}{6} + \frac{tan ^{8} ( x )}{8}) + C

x \to 6 lim (\frac{x - 6}{x ^{2} - x})

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + 2 y = 0

10 y^{^{''}} + 1000 y = 140 cos (8 t)

\int e^{2 x + 3} d x

What is the integral of 41tan^5(x)sec^4(x) ?
The integral of 41tan^5(x)sec^4(x) is 41((tan^6(x))/6+(tan^8(x))/8)+C