What is the integral of 2^{x-1}cos(2^x) ?

The integral of 2^{x-1}cos(2^x) is 1/(2ln(2))sin(2^x)+C

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integral of 2^{x-1}cos(2^x)

\int 2^{x - 1} cos (2^{x}) d x

\frac{1}{2 ln ( 2 )} sin (2^{x}) + C

Solution steps

\int 2^{x - 1} cos (2^{x}) d x

2^{x - 1} cos (2^{x}) = 2^{- 1} \cdot 2^{x} cos (2^{x})

= \int 2^{- 1} \cdot 2^{x} cos (2^{x}) d x

Take the constant out:

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

= 2^{- 1} \cdot \int 2^{x} cos (2^{x}) d x

Apply exponent rule:

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

= \frac{1}{2} \cdot \int 2^{x} cos (2^{x}) d x

Apply u-substitution

: \int \frac{cos ( u )}{ln ( 2 )} d u

= \frac{1}{2} \cdot \int \frac{cos ( u )}{ln ( 2 )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{ln ( 2 )} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{ln ( 2 )} sin (u)

Substitute back

u = 2^{x}

= \frac{1}{2} \cdot \frac{1}{ln ( 2 )} sin (2^{x})

Simplify

\frac{1}{2} \cdot \frac{1}{ln ( 2 )} sin (2^{x}) : \frac{1}{2 ln ( 2 )} sin (2^{x})

= \frac{1}{2 ln ( 2 )} sin (2^{x})

Add a constant to the solution

= \frac{1}{2 ln ( 2 )} sin (2^{x}) + C

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What is the integral of 2^{x-1}cos(2^x) ?
The integral of 2^{x-1}cos(2^x) is 1/(2ln(2))sin(2^x)+C