What is the integral of 2csc^4(x)cot^6(x) ?

The integral of 2csc^4(x)cot^6(x) is -2(1/7 cot^7(x)+1/9 cot^9(x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2csc^4(x)cot^6(x)

\int 2 csc^{4} (x) cot^{6} (x) d x

- 2 (\frac{1}{7} cot^{7} (x) + \frac{1}{9} cot^{9} (x)) + C

Solution steps

\int 2 csc^{4} (x) cot^{6} (x) d x

Take the constant out:

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

= 2 \cdot \int csc^{4} (x) cot^{6} (x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

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

= 2 (- \int 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

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

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

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

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

Substitute back

u = cot (x)

= 2 (- (\frac{cot ^{7} ( x )}{7} + \frac{cot ^{9} ( x )}{9}))

Simplify

2 (- (\frac{cot ^{7} ( x )}{7} + \frac{cot ^{9} ( x )}{9})) : - 2 (\frac{1}{7} cot^{7} (x) + \frac{1}{9} cot^{9} (x))

= - 2 (\frac{1}{7} cot^{7} (x) + \frac{1}{9} cot^{9} (x))

Add a constant to the solution

= - 2 (\frac{1}{7} cot^{7} (x) + \frac{1}{9} cot^{9} (x)) + C

x \to - 2 lim (7 - 2 x)

\int - 2 x e^{x^{2}} d x

\frac{d}{d x} (\frac{2}{ln ( 10 ) x ^{3}})

\int_{4}^{7} \frac{1}{x ^{2} - 16} d x

\frac{d}{d x} (2 x)

What is the integral of 2csc^4(x)cot^6(x) ?
The integral of 2csc^4(x)cot^6(x) is -2(1/7 cot^7(x)+1/9 cot^9(x))+C