What is the integral of 2xsec(3x^2)tan(3x^2) ?

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

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integral of 2xsec(3x^2)tan(3x^2)

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

\frac{1}{3} sec (3 x^{2}) + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int x sec (3 x^{2}) tan (3 x^{2}) d x

Apply u-substitution

: \int \frac{sec ( u ) tan ( u )}{6} d u

= 2 \cdot \int \frac{sec ( u ) tan ( u )}{6} d u

Take the constant out:

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

= 2 \cdot \frac{1}{6} \cdot \int sec (u) tan (u) d u

Apply u-substitution

: \int 1 d v

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

Integral of a constant:

\int a d x = a x

= 2 \cdot \frac{1}{6} \cdot 1 \cdot v

Substitute back

= 2 \cdot \frac{1}{6} \cdot 1 \cdot sec (3 x^{2})

Simplify

2 \cdot \frac{1}{6} \cdot 1 \cdot sec (3 x^{2}) : \frac{1}{3} sec (3 x^{2})

= \frac{1}{3} sec (3 x^{2})

Add a constant to the solution

= \frac{1}{3} sec (3 x^{2}) + C

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