integral of sin(2x)sin(nx)

解答

\int sin (2 x) sin (n x) d x

\frac{1}{2} (- \frac{1}{2 + n} sin (2 x + n x) + \frac{1}{2 - n} sin (2 x - n x)) + C

求解步骤

\int sin (2 x) sin (n x) d x

使用三角恒等式改写

= \int \frac{- cos ( 2 x + n x ) + cos ( 2 x - n 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 - cos (2 x + n x) + cos (2 x - n 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 cos (2 x + n x) d x + \int cos (2 x - n x) d x)

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

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

= \frac{1}{2} (- \frac{1}{2 + n} sin (2 x + n x) + \frac{1}{2 - n} sin (2 x - n x))

解答补常数

= \frac{1}{2} (- \frac{1}{2 + n} sin (2 x + n x) + \frac{1}{2 - n} sin (2 x - n x)) + C

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

\frac{d}{d x} (\frac{1}{1 - 2 x ^{2}})

s l o p e y = 6 x - x^{2}

\int e^{t} t d t

y^{^{'}} sin (x) + y cos (x) = x sin (x)