What is the integral of 4^tsin(4^t) ?

The integral of 4^tsin(4^t) is -(4^tcos(4^t))/(ln(2)*2^{2t+1)}+C

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integral of 4^tsin(4^t)

\int 4^{t} sin (4^{t}) d t

- \frac{4 ^{t} cos ( 4 ^{t} )}{ln ( 2 ) \cdot 2 ^{2 t + 1}} + C

Solution steps

\int 4^{t} sin (4^{t}) d t

Apply u-substitution

: \int \frac{u sin ( u )}{ln ( 2 ) \cdot 2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} d u

= \int \frac{u sin ( u )}{ln ( 2 ) \cdot 2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} d u

Take the constant out:

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

= \frac{1}{ln ( 2 )} \cdot \int \frac{u sin ( u )}{2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} d u

Apply Integration By Parts

: \frac{- u cos ( u ) + sin ( u )}{2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} - \int - \frac{- u cos ( u ) + sin ( u )}{2 u ^{2}} d u

= \frac{1}{ln ( 2 )} (\frac{- u cos ( u ) + sin ( u )}{2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} - \int - \frac{- u cos ( u ) + sin ( u )}{2 u ^{2}} d u)

\int - \frac{- u cos ( u ) + sin ( u )}{2 u ^{2}} d u = \frac{sin ( u )}{2 u}

= \frac{1}{ln ( 2 )} (\frac{- u cos ( u ) + sin ( u )}{2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} - \frac{sin ( u )}{2 u})

Substitute back

u = 4^{t}

= \frac{1}{ln ( 2 )} (\frac{- 4 ^{t} cos ( 4 ^{t} ) + sin ( 4 ^{t} )}{2 ^{\frac{l n ( 4 ^{t} ) + l n ( 2 )}{l n ( 2 )}}} - \frac{sin ( 4 ^{t} )}{2 \cdot 4 ^{t}})

Simplify

\frac{1}{ln ( 2 )} (\frac{- 4 ^{t} cos ( 4 ^{t} ) + sin ( 4 ^{t} )}{2 ^{\frac{l n ( 4 ^{t} ) + l n ( 2 )}{l n ( 2 )}}} - \frac{sin ( 4 ^{t} )}{2 \cdot 4 ^{t}}) : - \frac{4 ^{t} cos ( 4 ^{t} )}{ln ( 2 ) \cdot 2 ^{2 t + 1}}

= - \frac{4 ^{t} cos ( 4 ^{t} )}{ln ( 2 ) \cdot 2 ^{2 t + 1}}

Add a constant to the solution

= - \frac{4 ^{t} cos ( 4 ^{t} )}{ln ( 2 ) \cdot 2 ^{2 t + 1}} + C

\frac{d}{d x} (y - x + 1 + x - y + 1)

f (x) = ln (x + 9)

- 4 ln (+ \frac{1}{x} - 4)

(x^{2} + 2 y^{2}) d x = x y d y

\int \frac{sin ( \frac{1}{x ^{4}} )}{x ^{5}} d x

What is the integral of 4^tsin(4^t) ?
The integral of 4^tsin(4^t) is -(4^tcos(4^t))/(ln(2)*2^{2t+1)}+C