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

The integral of 1/(cos^2(x)+4cos^4(x)) is tan(x)-4/(sqrt(5))arctan(1/(sqrt(5))tan(x))+C

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

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

tan (x) - \frac{4}{5} arctan (\frac{1}{5} tan (x)) + C

Solution steps

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

Take the partial fraction of

\frac{1}{cos ^{2} ( x ) + 4 cos ^{4} ( x )} : \frac{1}{cos ^{2} ( x )} - \frac{4}{4 cos ^{2} ( x ) + 1}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int \frac{4}{4 cos ^{2} ( x ) + 1} d x = \frac{4}{5} arctan (\frac{1}{5} tan (x))

= tan (x) - \frac{4}{5} arctan (\frac{1}{5} tan (x))

Add a constant to the solution

= tan (x) - \frac{4}{5} arctan (\frac{1}{5} tan (x)) + C

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What is the integral of 1/(cos^2(x)+4cos^4(x)) ?
The integral of 1/(cos^2(x)+4cos^4(x)) is tan(x)-4/(sqrt(5))arctan(1/(sqrt(5))tan(x))+C