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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3cot^3(x)csc(x)

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

3 (- \frac{1}{3} csc^{3} (x) + csc (x)) + C

Solution steps

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

Take the constant out:

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

= 3 \cdot \int cot^{3} (x) csc (x) d x

Rewrite using trig identities

= 3 \cdot \int \frac{( 1 - sin ^{2} ( x ) ) cos ( x )}{sin ^{4} ( x )} d x

Apply u-substitution

: \int \frac{1 - u ^{2}}{u ^{4}} d u

= 3 \cdot \int \frac{1 - u ^{2}}{u ^{4}} d u

Expand

\frac{1 - u ^{2}}{u ^{4}} : \frac{1}{u ^{4}} - \frac{1}{u ^{2}}

= 3 \cdot \int \frac{1}{u ^{4}} - \frac{1}{u ^{2}} 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

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

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

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

= 3 (- \frac{1}{3 u ^{3}} - (- \frac{1}{u}))

Substitute back

u = sin (x)

= 3 (- \frac{1}{3 sin ^{3} ( x )} - (- \frac{1}{sin ( x )}))

Simplify

3 (- \frac{1}{3 sin ^{3} ( x )} - (- \frac{1}{sin ( x )})) : 3 (- \frac{1}{3} csc^{3} (x) + csc (x))

= 3 (- \frac{1}{3} csc^{3} (x) + csc (x))

Add a constant to the solution

= 3 (- \frac{1}{3} csc^{3} (x) + csc (x)) + C

\int \frac{3 x ^{2} + 7 x - 5}{3 x + 2} d x

\int \frac{1}{tan ( x ) ( 3 - 2 ln ( sin ( x ) ) )} d x

\int (2 x - 1) (- x^{2} + x)^{4} d x

\int - 7 x y^{4} d x

\int csc (0.5) d x

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