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

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

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integral of 1/(tan^2(x))

integral

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

- x - cot (x) + C

Solution steps

\int \frac{1}{tan ^{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

in t e g r a l \frac{x ^{3}}{( 1 + x ^{2} ) ^{2}}

in t e g r a l sin^{2} (x) + cos^{2} (x)

in t e g r a l \frac{ln ( x )}{( x + 1 ) ^{2}}

in t e g r a l \frac{sin ^{2} ( x )}{2}

in t e g r a l \frac{1}{( e ^{x} + e ^{- x} )}

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