What is the integral of sec^2(3x)e^{tan(3x)} ?

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

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integral of sec^2(3x)e^{tan(3x)}

\int sec^{2} (3 x) e^{t a n (3 x)} d x

\frac{1}{3} e^{t a n (3 x)} + C

Solution steps

\int sec^{2} (3 x) e^{t a n (3 x)} d x

Apply u-substitution

: \int sec^{2} (u) e^{t a n (u)} \frac{1}{3} d u

= \int sec^{2} (u) e^{t a n (u)} \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sec^{2} (u) e^{t a n (u)} d u

Apply u-substitution

: \int e^{v} d v

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

Use the common integral:

\int e^{v} d v = e^{v}

= \frac{1}{3} e^{v}

Substitute back

= \frac{1}{3} e^{t a n (3 x)}

Add a constant to the solution

= \frac{1}{3} e^{t a n (3 x)} + C

y^{^{'}} - y = x

x \to 0 lim (3 - x^{2})

\frac{d}{d s} (\frac{e ^{s}}{s})

f (x) = (1 - 2 x) (x - 1)^{3}

x \to 2 - lim (csc (\frac{π x}{2}))

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