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

The integral of 1/(1+cos^2(θ)) is 1/(sqrt(2))arctan((tan(θ))/(sqrt(2)))+C

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integral of 1/(1+cos^2(θ))

\int \frac{1}{1 + cos ^{2} ( θ )} d θ

\frac{1}{2} arctan (\frac{tan ( θ )}{2}) + C

Solution steps

\int \frac{1}{1 + cos ^{2} ( θ )} d θ

Rewrite using trig identities

= \int \frac{2}{cos ( 2 θ ) + 3} d θ

Take the constant out:

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

= 2 \cdot \int \frac{1}{cos ( 2 θ ) + 3} d θ

Apply u-substitution

: \int \frac{1}{2 ( cos ( u ) + 3 )} d u

= 2 \cdot \int \frac{1}{2 ( cos ( u ) + 3 )} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{cos ( u ) + 3} d u

Apply u-substitution

: \int \frac{1}{v ^{2} + 2} d v

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{2} + 2} d v

Apply Integral Substitution

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{2 ( w ^{2} + 1 )} d w

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} arctan (w)

Substitute back

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} arctan (\frac{tan ( \frac{2 θ}{2} )}{2})

Simplify

2 \cdot \frac{1}{2} \cdot \frac{1}{2} arctan (\frac{tan ( \frac{2 θ}{2} )}{2}) : \frac{1}{2} arctan (\frac{tan ( θ )}{2})

= \frac{1}{2} arctan (\frac{tan ( θ )}{2})

Add a constant to the solution

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

\frac{d}{d x} (arccos (sin (x^{2} + 9 x)))

f (x) = 3 x + 4

x \to 0 lim ((1 + \frac{3}{x})^{x})

f (x) = \frac{2}{4 - x}, at x = 1

\int_{0}^{1} cos (\frac{π t}{2}) d t

What is the integral of 1/(1+cos^2(θ)) ?
The integral of 1/(1+cos^2(θ)) is 1/(sqrt(2))arctan((tan(θ))/(sqrt(2)))+C