What is the integral of (sec^2(x))/(1+9tan^2(x)) ?

The integral of (sec^2(x))/(1+9tan^2(x)) is 1/3 arctan(3tan(x))+C

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integral of (sec^2(x))/(1+9tan^2(x))

\int \frac{sec ^{2} ( x )}{1 + 9 tan ^{2} ( x )} d x

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

Solution steps

\int \frac{sec ^{2} ( x )}{1 + 9 tan ^{2} ( x )} d x

Apply u-substitution

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{3} arctan (v)

Substitute back

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

Add a constant to the solution

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

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