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

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

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

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

\frac{1}{3} e^{t a n (3 x) - 2} + C

Solution steps

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

sec^{2} (3 x) e^{t a n (3 x) - 2} = e^{- 2} sec^{2} (3 x) e^{t a n (3 x)}

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

Take the constant out:

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

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

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

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

Apply u-substitution

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

Apply u-substitution

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

Use the common integral:

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

= \frac{1}{e ^{2}} \cdot \frac{1}{3} e^{v}

Substitute back

= \frac{1}{e ^{2}} \cdot \frac{1}{3} e^{t a n (3 x)}

Simplify

\frac{1}{e ^{2}} \cdot \frac{1}{3} e^{t a n (3 x)} : \frac{1}{3} e^{t a n (3 x) - 2}

= \frac{1}{3} e^{t a n (3 x) - 2}

Add a constant to the solution

= \frac{1}{3} e^{t a n (3 x) - 2} + C

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