What is the integral of 6cot^3(x)csc^3(x) ?

The integral of 6cot^3(x)csc^3(x) is -6(1/5 csc^5(x)-1/3 csc^3(x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 6cot^3(x)csc^3(x)

\int 6 cot^{3} (x) csc^{3} (x) d x

- 6 (\frac{1}{5} csc^{5} (x) - \frac{1}{3} csc^{3} (x)) + C

Solution steps

\int 6 cot^{3} (x) csc^{3} (x) d x

Take the constant out:

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

= 6 \cdot \int cot^{3} (x) csc^{3} (x) d x

Apply u-substitution

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

= 6 \cdot \int - u^{2} (u^{2} - 1) d u

Take the constant out:

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

= 6 (- \int u^{2} (u^{2} - 1) d u)

Expand

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

= 6 (- \int u^{4} - u^{2} 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

= 6 (- (\int u^{4} d u - \int u^{2} d u))

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

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

= 6 (- (\frac{u ^{5}}{5} - \frac{u ^{3}}{3}))

Substitute back

u = csc (x)

= 6 (- (\frac{csc ^{5} ( x )}{5} - \frac{csc ^{3} ( x )}{3}))

Simplify

6 (- (\frac{csc ^{5} ( x )}{5} - \frac{csc ^{3} ( x )}{3})) : - 6 (\frac{1}{5} csc^{5} (x) - \frac{1}{3} csc^{3} (x))

= - 6 (\frac{1}{5} csc^{5} (x) - \frac{1}{3} csc^{3} (x))

Add a constant to the solution

= - 6 (\frac{1}{5} csc^{5} (x) - \frac{1}{3} csc^{3} (x)) + C

\int \frac{1}{sec ( x ) ( 1 + cos ( x ) )} d x

x 4 + y 3 x 5 + 2 y^{^{'}} = 0

\frac{d}{d x} (e^{3 x} + b)

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

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

What is the integral of 6cot^3(x)csc^3(x) ?
The integral of 6cot^3(x)csc^3(x) is -6(1/5 csc^5(x)-1/3 csc^3(x))+C