integral of 4^tsin(4^t)

解答

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

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

求解步骤

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

使用换元积分法

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

提出常数:

\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

使用分布积分法

= \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})

u = 4^{t}

代回

= \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}})

化简

\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}}

解答补常数

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

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

d er i v a t i v e f (x) = ln (x + 9)

d er i v a t i v e - 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