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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply Trigonometric Substitution

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

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

Apply Integral Substitution

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

Substitute back

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

Add a constant to the solution

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

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