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

The integral of 1/(2-sin^2(x)) is 1/(sqrt(2))arctan((tan(x))/(sqrt(2)))+C

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integral of 1/(2-sin^2(x))

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

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

Solution steps

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

Rewrite using trig identities

= \int \frac{2}{cos ( 2 x ) + 3} d x

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 x ) + 3} d x

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 x}{2} )}{2})

Simplify

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

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

Add a constant to the solution

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

(0) (4.2)

\int \frac{- 5}{( x + 2 ) ^{2}} d x

\int - x^{2} + 2 d x

t \frac{d x}{d t} + 2 x = 3 t

\frac{10}{x ^{2} + 1}

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