What is the integral of (3^{x-1})/(sec(3^x)) ?

The integral of (3^{x-1})/(sec(3^x)) is 1/(3ln(3))sin(3^x)+C

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integral of (3^{x-1})/(sec(3^x))

\int \frac{3 ^{x - 1}}{sec ( 3 ^{x} )} d x

\frac{1}{3 ln ( 3 )} sin (3^{x}) + C

Solution steps

\int \frac{3 ^{x - 1}}{sec ( 3 ^{x} )} d x

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

= \int 3^{- 1} \cdot \frac{3 ^{x}}{sec ( 3 ^{x} )} d x

Take the constant out:

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

= 3^{- 1} \cdot \int \frac{3 ^{x}}{sec ( 3 ^{x} )} d x

Apply exponent rule:

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

= \frac{1}{3} \cdot \int \frac{3 ^{x}}{sec ( 3 ^{x} )} d x

Use the basic trigonometric identity:

\frac{1}{sec ( x )} = cos (x)

= \frac{1}{3} \cdot \int 3^{x} cos (3^{x}) d x

Apply u-substitution

: \int \frac{cos ( u )}{ln ( 3 )} d u

= \frac{1}{3} \cdot \int \frac{cos ( u )}{ln ( 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 \frac{1}{ln ( 3 )} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{3} \cdot \frac{1}{ln ( 3 )} sin (u)

Substitute back

u = 3^{x}

= \frac{1}{3} \cdot \frac{1}{ln ( 3 )} sin (3^{x})

Simplify

\frac{1}{3} \cdot \frac{1}{ln ( 3 )} sin (3^{x}) : \frac{1}{3 ln ( 3 )} sin (3^{x})

= \frac{1}{3 ln ( 3 )} sin (3^{x})

Add a constant to the solution

= \frac{1}{3 ln ( 3 )} sin (3^{x}) + C

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