What is the integral of 1/(4sin^2(x)+9cos^2(x)) ?

The integral of 1/(4sin^2(x)+9cos^2(x)) is 1/6 arctan(2/3 tan(x))+C

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integral of 1/(4sin^2(x)+9cos^2(x))

\int \frac{1}{4 sin ^{2} ( x ) + 9 cos ^{2} ( x )} d x

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

Solution steps

\int \frac{1}{4 sin ^{2} ( x ) + 9 cos ^{2} ( x )} d x

Apply Trigonometric Substitution

: \int \frac{1}{4 u ^{2} + 9} d u

= \int \frac{1}{4 u ^{2} + 9} d u

Apply Integral Substitution

= \int \frac{1}{6 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{6} arctan (v)

Substitute back

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

Add a constant to the solution

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

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What is the integral of 1/(4sin^2(x)+9cos^2(x)) ?
The integral of 1/(4sin^2(x)+9cos^2(x)) is 1/6 arctan(2/3 tan(x))+C