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

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

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integral of 2csc^3(x)

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

\frac{1}{4} (- cot^{2} (\frac{x}{2}) + 4 ln tan (\frac{x}{2}) + tan^{2} (\frac{x}{2})) + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int csc^{3} (x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

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

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

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

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

\int \frac{2}{u} d u = 2 ln ∣ u ∣

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

= 2 \cdot \frac{1}{4} (- \frac{1}{2 u ^{2}} + 2 ln ∣ u ∣ + \frac{u ^{2}}{2})

Substitute back

u = tan (\frac{x}{2})

= 2 \cdot \frac{1}{4} (- \frac{1}{2 tan ^{2} ( \frac{x}{2} )} + 2 ln tan (\frac{x}{2}) + \frac{tan ^{2} ( \frac{x}{2} )}{2})

Simplify

2 \cdot \frac{1}{4} (- \frac{1}{2 tan ^{2} ( \frac{x}{2} )} + 2 ln tan (\frac{x}{2}) + \frac{tan ^{2} ( \frac{x}{2} )}{2}) : \frac{1}{4} (- cot^{2} (\frac{x}{2}) + 4 ln tan (\frac{x}{2}) + tan^{2} (\frac{x}{2}))

= \frac{1}{4} (- cot^{2} (\frac{x}{2}) + 4 ln tan (\frac{x}{2}) + tan^{2} (\frac{x}{2}))

Add a constant to the solution

= \frac{1}{4} (- cot^{2} (\frac{x}{2}) + 4 ln tan (\frac{x}{2}) + tan^{2} (\frac{x}{2})) + C

(1 - x)^{2} \cdot b

\frac{d}{d x} (2 x^{- 4})

\int x^{3} ln (5 x) d x

(x^{2} e^{x})^{^{'}}

y^{^{'}} + \frac{2 y}{x} = - \frac{4}{x}

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