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

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

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integral of 1+cot^2(x)

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

- cot (x) + C

Solution steps

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

Rewrite using trig identities

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

Use the common integral:

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

= - cot (x)

Add a constant to the solution

= - cot (x) + C

\frac{d}{d x} (\frac{x ^{3}}{6} - 3 x)

\int \frac{1}{36} x d x

\int_{0}^{4} \frac{1}{x ^{2} - 2 x - 3} d x

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

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