What is the integral of-10csc^2(5x)cot^3(5x) ?

The integral of-10csc^2(5x)cot^3(5x) is 1/2 cot^4(5x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of-10csc^2(5x)cot^3(5x)

\int - 10 csc^{2} (5 x) cot^{3} (5 x) d x

\frac{1}{2} cot^{4} (5 x) + C

Solution steps

\int - 10 csc^{2} (5 x) cot^{3} (5 x) d x

Take the constant out:

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

= - 10 \cdot \int csc^{2} (5 x) cot^{3} (5 x) d x

Apply u-substitution

: \int csc^{2} (u) cot^{3} (u) \frac{1}{5} d u

= - 10 \cdot \int csc^{2} (u) cot^{3} (u) \frac{1}{5} d u

Take the constant out:

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

= - 10 \cdot \frac{1}{5} \cdot \int csc^{2} (u) cot^{3} (u) d u

Apply u-substitution

: \int - v^{3} d v

= - 10 \cdot \frac{1}{5} \cdot \int - v^{3} d v

Take the constant out:

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

= - 10 \cdot \frac{1}{5} (- \int v^{3} d v)

Apply the Power Rule

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

= - 10 \cdot \frac{1}{5} (- \frac{v ^{4}}{4})

Substitute back

= - 10 \cdot \frac{1}{5} (- \frac{cot ^{4} ( 5 x )}{4})

Simplify

- 10 \cdot \frac{1}{5} (- \frac{cot ^{4} ( 5 x )}{4}) : \frac{1}{2} cot^{4} (5 x)

= \frac{1}{2} cot^{4} (5 x)

Add a constant to the solution

= \frac{1}{2} cot^{4} (5 x) + C

\int x 16 - x d x

\int \frac{3 x}{- 2} d x

\int \frac{y ^{2}}{( x ^{2} + y ^{2} ) ^{\frac{3}{2}}} d x

\int 3 x - \frac{1}{2 x} d x

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

What is the integral of-10csc^2(5x)cot^3(5x) ?
The integral of-10csc^2(5x)cot^3(5x) is 1/2 cot^4(5x)+C