What is the integral of (sec^2(2x))/(1+3tan^2(2x)) ?

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

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integral of (sec^2(2x))/(1+3tan^2(2x))

\int \frac{sec ^{2} ( 2 x )}{1 + 3 tan ^{2} ( 2 x )} d x

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

Solution steps

\int \frac{sec ^{2} ( 2 x )}{1 + 3 tan ^{2} ( 2 x )} d x

Apply u-substitution

: \int \frac{sec ^{2} ( u )}{2 ( 1 + 3 tan ^{2} ( u ) )} d u

= \int \frac{sec ^{2} ( u )}{2 ( 1 + 3 tan ^{2} ( 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{sec ^{2} ( u )}{1 + 3 tan ^{2} ( u )} d u

Apply u-substitution

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

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

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{3 ( 1 + w ^{2} )} 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}{3} \cdot \int \frac{1}{1 + w ^{2}} d w

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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