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

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

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integral of 1/(sec^2(θ))

\int \frac{1}{sec ^{2} ( θ )} d θ

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

Solution steps

\int \frac{1}{sec ^{2} ( θ )} d θ

Rewrite using trig identities

= \int \frac{1 + cos ( 2 θ )}{2} d θ

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 θ) d θ

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 θ + \int cos (2 θ) d θ)

\int 1 d θ = θ

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

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

Add a constant to the solution

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

\frac{d ^{2}}{d x ^{2}} (ln (1 - x))

\int sin (x) d (sin (x)) d x

y = 4 x e^{x}, (0, 0)

\int x (cos (x)) d x

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