What is the integral of cot^3(4x)csc^2(4x) ?

The integral of cot^3(4x)csc^2(4x) is -1/16 cot^4(4x)+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(4x)csc^2(4x)

\int cot^{3} (4 x) csc^{2} (4 x) d x

- \frac{1}{16} cot^{4} (4 x) + C

Solution steps

\int cot^{3} (4 x) csc^{2} (4 x) d x

Apply u-substitution

: \int cot^{3} (u) csc^{2} (u) \frac{1}{4} d u

= \int cot^{3} (u) csc^{2} (u) \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int cot^{3} (u) csc^{2} (u) d u

Apply u-substitution

: \int - v^{3} d v

= \frac{1}{4} \cdot \int - v^{3} d v

Take the constant out:

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

= \frac{1}{4} (- \int v^{3} d v)

Apply the Power Rule

: \frac{v ^{4}}{4}

= \frac{1}{4} (- \frac{v ^{4}}{4})

Substitute back

= \frac{1}{4} (- \frac{cot ^{4} ( 4 x )}{4})

Simplify

\frac{1}{4} (- \frac{cot ^{4} ( 4 x )}{4}) : - \frac{1}{16} cot^{4} (4 x)

= - \frac{1}{16} cot^{4} (4 x)

Add a constant to the solution

= - \frac{1}{16} cot^{4} (4 x) + C

\int (x + 1)^{2} ln (3 x) d x

\int (e^{x} + \frac{1}{x ^{2}}) d x

\int \frac{1}{u ^{4} + u} d u

\int \frac{x ^{3} + x - 4}{x - 1} d x

What is the integral of cot^3(4x)csc^2(4x) ?
The integral of cot^3(4x)csc^2(4x) is -1/16 cot^4(4x)+C