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

The integral of 1/(2+sin^2(x)) is 1/(sqrt(6))arctan((sqrt(3))/(sqrt(2))tan(x))+C

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

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

\frac{1}{6} arctan (\frac{3}{2} tan (x)) + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{1}{2 ( - cos ( u ) + 5 )} d u

= 2 \cdot \int \frac{1}{2 ( - cos ( u ) + 5 )} 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 ) + 5} d u

Apply u-substitution

: \int \frac{1}{3 v ^{2} + 2} d v

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{3 v ^{2} + 2} d v

Apply Integral Substitution

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{6 ( 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}{6} \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}{6} arctan (w)

Substitute back

= 2 \cdot \frac{1}{2} \cdot \frac{1}{6} arctan (\frac{3}{2} tan (\frac{2 x}{2}))

Simplify

2 \cdot \frac{1}{2} \cdot \frac{1}{6} arctan (\frac{3}{2} tan (\frac{2 x}{2})) : \frac{1}{6} arctan (\frac{3}{2} tan (x))

= \frac{1}{6} arctan (\frac{3}{2} tan (x))

Add a constant to the solution

= \frac{1}{6} arctan (\frac{3}{2} tan (x)) + C

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What is the integral of 1/(2+sin^2(x)) ?
The integral of 1/(2+sin^2(x)) is 1/(sqrt(6))arctan((sqrt(3))/(sqrt(2))tan(x))+C