What is the integral of 3/(2sin^2(x)+5) ?

The integral of 3/(2sin^2(x)+5) is 3/(sqrt(35))arctan((sqrt(7))/(sqrt(5))tan(x))+C

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integral of 3/(2sin^2(x)+5)

\int \frac{3}{2 sin ^{2} ( x ) + 5} d x

\frac{3}{35} arctan (\frac{7}{5} tan (x)) + C

Solution steps

\int \frac{3}{2 sin ^{2} ( x ) + 5} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{2 sin ^{2} ( x ) + 5} d x

Rewrite using trig identities

= 3 \cdot \int \frac{1}{- cos ( 2 x ) + 6} d x

Apply u-substitution

: \int \frac{1}{2 ( - cos ( u ) + 6 )} d u

= 3 \cdot \int \frac{1}{2 ( - cos ( u ) + 6 )} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{- cos ( u ) + 6} d u

Apply u-substitution

: \int \frac{2}{7 v ^{2} + 5} d v

= 3 \cdot \frac{1}{2} \cdot \int \frac{2}{7 v ^{2} + 5} d v

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot 2 \cdot \int \frac{1}{7 v ^{2} + 5} d v

Apply Integral Substitution

= 3 \cdot \frac{1}{2} \cdot 2 \cdot \int \frac{1}{35 ( w ^{2} + 1 )} d w

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot 2 \cdot \frac{1}{35} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= 3 \cdot \frac{1}{2} \cdot 2 \cdot \frac{1}{35} arctan (w)

Substitute back

= 3 \cdot \frac{1}{2} \cdot 2 \cdot \frac{1}{35} arctan (\frac{7}{5} tan (\frac{2 x}{2}))

Simplify

3 \cdot \frac{1}{2} \cdot 2 \cdot \frac{1}{35} arctan (\frac{7}{5} tan (\frac{2 x}{2})) : \frac{3}{35} arctan (\frac{7}{5} tan (x))

= \frac{3}{35} arctan (\frac{7}{5} tan (x))

Add a constant to the solution

= \frac{3}{35} arctan (\frac{7}{5} tan (x)) + C

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What is the integral of 3/(2sin^2(x)+5) ?
The integral of 3/(2sin^2(x)+5) is 3/(sqrt(35))arctan((sqrt(7))/(sqrt(5))tan(x))+C