What is the integral of 1/2 cot^2(x)csc^4(x) ?

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

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integral of 1/2 cot^2(x)csc^4(x)

\int \frac{1}{2} cot^{2} (x) csc^{4} (x) d x

- \frac{1}{2} (\frac{1}{5} cot^{5} (x) + \frac{1}{3} cot^{3} (x)) + C

Solution steps

\int \frac{1}{2} cot^{2} (x) csc^{4} (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^{2} (x) csc^{4} (x) d x

Apply Trigonometric Substitution

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

Take the constant out:

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

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

Expand

u^{2} (u^{2} + 1) : u^{4} + u^{2}

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

\int u^{4} d u = \frac{u ^{5}}{5}

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

= \frac{1}{2} (- (\frac{u ^{5}}{5} + \frac{u ^{3}}{3}))

Substitute back

u = cot (x)

= \frac{1}{2} (- (\frac{cot ^{5} ( x )}{5} + \frac{cot ^{3} ( x )}{3}))

Simplify

\frac{1}{2} (- (\frac{cot ^{5} ( x )}{5} + \frac{cot ^{3} ( x )}{3})) : - \frac{1}{2} (\frac{1}{5} cot^{5} (x) + \frac{1}{3} cot^{3} (x))

= - \frac{1}{2} (\frac{1}{5} cot^{5} (x) + \frac{1}{3} cot^{3} (x))

Add a constant to the solution

= - \frac{1}{2} (\frac{1}{5} cot^{5} (x) + \frac{1}{3} cot^{3} (x)) + C

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