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

The integral of 1/(tan^2(u)+1) is 1/2 (u+1/2 sin(2u))+C

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integral of 1/(tan^2(u)+1)

\int \frac{1}{tan ^{2} ( u ) + 1} d u

\frac{1}{2} (u + \frac{1}{2} sin (2 u)) + C

Solution steps

\int \frac{1}{tan ^{2} ( u ) + 1} d u

Rewrite using trig identities

= \int \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int 1 + cos (2 u) d u

Apply the Sum Rule:

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

= \frac{1}{2} (\int 1 d u + \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Add a constant to the solution

= \frac{1}{2} (u + \frac{1}{2} sin (2 u)) + C

x \to 1 - lim (\frac{1}{x ^{2} - 1})

\int \frac{2}{3 x} d x

x \to \frac{π}{2} lim (csc (x))

y = x - 1, (10, 3)

f (x) = 2 e^{0}

What is the integral of 1/(tan^2(u)+1) ?
The integral of 1/(tan^2(u)+1) is 1/2 (u+1/2 sin(2u))+C