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

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

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integral of 1-cot^2(x)

integral

1 - cot^{2} (x)

2 x + cot (x) + C

Solution steps

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

\int 1 d x = x

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

= x - (- x - cot (x))

Simplify

= 2 x + cot (x)

Add a constant to the solution

= 2 x + cot (x) + C

\frac{- 2}{x ^{3}}

\frac{ln ( x ^{2} )}{x ^{3}}

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

\frac{x}{sin ^{2} ( x )}

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