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

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

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

integral

cot^{4} (x) - cot^{2} (x)

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

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int cot^{4} (x) d x = - \frac{1}{3} cot^{3} (x) + x + cot (x)

\int cot^{2} (x) d x = - x - cot (x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{6}{x ^{3}}

y = x^{2} + x - 12

\frac{x + 1}{( x ^{2} + 4 x + 8 )}

\frac{u}{( 1 + u )}

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

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