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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- 2 cot (2 x) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{8}{1 - cos ( 4 x )} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{1 - cos ( 4 x )} d x

Apply u-substitution

: \int \frac{1}{4 ( 1 - cos ( u ) )} d u

= 8 \cdot \int \frac{1}{4 ( 1 - cos ( u ) )} d u

Take the constant out:

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

= 8 \cdot \frac{1}{4} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= 8 \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 8 \cdot \frac{1}{4} (- \frac{1}{v})

Substitute back

= 8 \cdot \frac{1}{4} (- \frac{1}{tan ( \frac{4 x}{2} )})

Simplify

8 \cdot \frac{1}{4} (- \frac{1}{tan ( \frac{4 x}{2} )}) : - 2 cot (2 x)

= - 2 cot (2 x)

Add a constant to the solution

= - 2 cot (2 x) + C

\int x^{2} - 4 x + 4 d x

\frac{d y}{d x} = \frac{cos ( 12 x )}{e ^{12 y}}

\int_{1}^{2} \frac{15}{x ^{3} + 5 x} d x

xL (x) - x

y^{^{''}} - y^{^{'}} - 6 y = 0, y (0) = 0, y^{^{'}} (0) = 4

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