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

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

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integral of 1/(sec^2(u))

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

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

Solution steps

\int \frac{1}{sec ^{2} ( u )} 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

\int_{0}^{4} ∣ 2 x - 3 ∣ d x

\frac{s}{s + \frac{1}{2}}

e^{- x}, 5

\int 1 + \frac{2}{y} d y

\frac{1}{2 z ^{3}}

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