What is the integral of 1/(3+sin^2(x)-cos^2(x)) ?

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

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integral of 1/(3+sin^2(x)-cos^2(x))

\int \frac{1}{3 + sin ^{2} ( x ) - cos ^{2} ( x )} d x

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

Solution steps

\int \frac{1}{3 + sin ^{2} ( x ) - cos ^{2} ( x )} d x

Simplify

\frac{1}{3 + sin ^{2} ( x ) - cos ^{2} ( x )} : \frac{1}{3 - cos ( 2 x )}

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

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

Apply Integral Substitution

= \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

= \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)

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

h (x) = \frac{1}{x ^{2} + 4 x - 21}

\int (x + \frac{1}{x ^{4}}) d x

\frac{d}{d x} (19)

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

7 x, x \cdot (1 5^{2} - x^{2})

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