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

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

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integral of cot^2(x)(1)

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

- x - cot (x) + C

Solution steps

\int cot^{2} (x) \cdot 1 d x

Simplify

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

Rewrite using trig identities

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

\int 1 d x = x

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

= - x - cot (x)

Add a constant to the solution

= - x - cot (x) + C

\int sin (4 x) sin (n x) d x

y^{^{''}} - 3 y^{^{'}} - 10 y = 0, y (0) = 1, y^{^{'}} (0) = 10

x \to \infty lim ((0.6)^{\frac{1}{x}})

\int 2 x (x^{2} + 4)^{4} d x

\int 3 x 1 - x^{4} d x

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