integral of sin(2x)cos(3x)

解答

\int sin (2 x) cos (3 x) d x

\frac{1}{2} (- \frac{1}{5} cos (5 x) + cos (x)) + C

求解步骤

\int sin (2 x) cos (3 x) d x

利用以下特性:

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

= \int \frac{sin ( 2 x + 3 x ) + sin ( 2 x - 3 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 (2 x + 3 x) + sin (2 x - 3 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 (2 x + 3 x) d x + \int sin (2 x - 3 x) d x)

\int sin (2 x + 3 x) d x = - \frac{1}{5} cos (5 x)

\int sin (2 x - 3 x) d x = cos (x)

= \frac{1}{2} (- \frac{1}{5} cos (5 x) + cos (x))

解答补常数

= \frac{1}{2} (- \frac{1}{5} cos (5 x) + cos (x)) + C

\int sec^{2} (3 x) e^{t a n (3 x)} d x

x \to 0 lim (3 - x^{2})

x \to 2 - lim (csc (\frac{π x}{2}))

y^{^{''}} + y = 1

ma c l a u r in f (x) = sin (x)