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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

= \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

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

Expand

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

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

Substitute back

u = cot (x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (sin (3 x + 2 y))

\frac{s}{( s ^{2} + 16 ) ^{2}}

4 y^{^{''}} + 6 y^{^{'}} + 7 y = 0

\frac{d}{d x} ((4 x - x^{2})^{100})

\frac{d}{d x} (sin (e^{2 x y}) + e^{s i n (2 x y)})

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