What is the integral of cot^{-3}(3y)csc^4(3y) ?

The integral of cot^{-3}(3y)csc^4(3y) is 1/9 (3/2 tan^2(3y)+ln|tan^3(3y)|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of cot^{-3}(3y)csc^4(3y)

\int cot^{- 3} (3 y) csc^{4} (3 y) d y

\frac{1}{9} (\frac{3}{2} tan^{2} (3 y) + ln tan^{3} (3 y)) + C

Solution steps

\int cot^{- 3} (3 y) csc^{4} (3 y) d y

Apply u-substitution

: \int \frac{u ^{\frac{2}{3}} + 1}{9 u} d u

= \int \frac{u ^{\frac{2}{3}} + 1}{9 u} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{u ^{\frac{2}{3}} + 1}{u} d u

Expand

\frac{u ^{\frac{2}{3}} + 1}{u} : \frac{1}{u ^{\frac{1}{3}}} + \frac{1}{u}

= \frac{1}{9} \cdot \int \frac{1}{u ^{\frac{1}{3}}} + \frac{1}{u} 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

= \frac{1}{9} (\int \frac{1}{u ^{\frac{1}{3}}} d u + \int \frac{1}{u} d u)

\int \frac{1}{u ^{\frac{1}{3}}} d u = \frac{3}{2} u^{\frac{2}{3}}

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{9} (\frac{3}{2} u^{\frac{2}{3}} + ln ∣ u ∣)

Substitute back

u = cot^{- 3} (3 y)

= \frac{1}{9} (\frac{3}{2} (cot^{- 3} (3 y))^{\frac{2}{3}} + ln cot^{- 3} (3 y))

Simplify

\frac{1}{9} (\frac{3}{2} (cot^{- 3} (3 y))^{\frac{2}{3}} + ln cot^{- 3} (3 y)) : \frac{1}{9} (\frac{3}{2} tan^{2} (3 y) + ln tan^{3} (3 y))

= \frac{1}{9} (\frac{3}{2} tan^{2} (3 y) + ln tan^{3} (3 y))

Add a constant to the solution

= \frac{1}{9} (\frac{3}{2} tan^{2} (3 y) + ln tan^{3} (3 y)) + C

\int 12 x - 12 x^{3} d x

\int (x + 1)^{10} (x + 5) d x

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

\int \frac{x + 10}{x ^{2} + 3 x + 2} d x

\int 2 cos (2 x) - 6 d x

What is the integral of cot^{-3}(3y)csc^4(3y) ?
The integral of cot^{-3}(3y)csc^4(3y) is 1/9 (3/2 tan^2(3y)+ln|tan^3(3y)|)+C