integral of sin(6x)cos(4x)

解答

\int sin (6 x) cos (4 x) d x

\frac{1}{2} (- \frac{1}{10} cos (10 x) - \frac{1}{2} cos (2 x)) + C

求解步骤

\int sin (6 x) cos (4 x) d x

利用以下特性:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

= \int \frac{sin ( 6 x + 4 x ) + sin ( 6 x - 4 x )}{2} d x

提出常数:

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

= \frac{1}{2} \cdot \int sin (6 x + 4 x) + sin (6 x - 4 x) d x

化简

= \frac{1}{2} \cdot \int sin (10 x) + sin (2 x) d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int sin (10 x) d x + \int sin (2 x) d x)

\int sin (10 x) d x = - \frac{1}{10} cos (10 x)

\int sin (2 x) d x = - \frac{1}{2} cos (2 x)

= \frac{1}{2} (- \frac{1}{10} cos (10 x) - \frac{1}{2} cos (2 x))

解答补常数

= \frac{1}{2} (- \frac{1}{10} cos (10 x) - \frac{1}{2} cos (2 x)) + C

t \to 0 lim (\frac{e ^{2 t}}{sin ( t )})

\int \frac{( x + 1 )}{7 x + x ^{2}} d x

\frac{d N}{d t} = N (1 - 0.001 N)

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

\frac{d}{d x} (\frac{1}{3} x^{3} + \frac{1}{2} x^{2} - 6 x + 8)