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

The integral of cot^4(x)csc^4(x) is -1/7 cot^7(x)-1/5 cot^5(x)+C

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

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

- \frac{1}{7} cot^{7} (x) - \frac{1}{5} cot^{5} (x) + C

Solution steps

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

Apply Trigonometric Substitution

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

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

Expand

u^{4} (u^{2} + 1) : u^{6} + u^{4}

= - \int u^{6} + u^{4} 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^{6} d u + \int u^{4} d u)

\int u^{6} d u = \frac{u ^{7}}{7}

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

= - (\frac{u ^{7}}{7} + \frac{u ^{5}}{5})

Substitute back

u = cot (x)

= - (\frac{cot ^{7} ( x )}{7} + \frac{cot ^{5} ( x )}{5})

Simplify

- (\frac{cot ^{7} ( x )}{7} + \frac{cot ^{5} ( x )}{5}) : - \frac{1}{7} cot^{7} (x) - \frac{1}{5} cot^{5} (x)

= - \frac{1}{7} cot^{7} (x) - \frac{1}{5} cot^{5} (x)

Add a constant to the solution

= - \frac{1}{7} cot^{7} (x) - \frac{1}{5} cot^{5} (x) + C

\int_{0}^{\frac{π}{2}} 5 cos^{2} (x) d x

y^{^{''}} - 4 y^{^{'}} + 16 y = 0

\frac{\partial}{\partial x} (\frac{1}{x + 2 c t})

x \to 0 lim (\frac{x + 2}{x ^{3} + 8})

lo g_{3} (3 x^{3} + 4 x + 4)

What is the integral of cot^4(x)csc^4(x) ?
The integral of cot^4(x)csc^4(x) is -1/7 cot^7(x)-1/5 cot^5(x)+C