What is the integral of 1/2 cot^3(x)csc(x) ?

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

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integral of 1/2 cot^3(x)csc(x)

\int \frac{1}{2} cot^{3} (x) csc (x) d x

\frac{1}{2} (- \frac{1}{3} csc^{3} (x) + csc (x)) + C

Solution steps

\int \frac{1}{2} 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

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

Rewrite using trig identities

= \frac{1}{2} \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

= \frac{1}{2} \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}}

= \frac{1}{2} \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

= \frac{1}{2} (\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}

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

Substitute back

u = sin (x)

= \frac{1}{2} (- \frac{1}{3 sin ^{3} ( x )} - (- \frac{1}{sin ( x )}))

Simplify

\frac{1}{2} (- \frac{1}{3 sin ^{3} ( x )} - (- \frac{1}{sin ( x )})) : \frac{1}{2} (- \frac{1}{3} csc^{3} (x) + csc (x))

= \frac{1}{2} (- \frac{1}{3} csc^{3} (x) + csc (x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{3} csc^{3} (x) + csc (x)) + C

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