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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Substitution

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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