integral of cos(x)cos(2nx)

解答

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

\frac{1}{2} (\frac{1}{n} sin (2 n x) cos (x) + \frac{1}{4 n} (- \frac{2}{2 n + 1} sin (2 n x + x) + \frac{2}{2 n - 1} sin (2 n x - x))) + C

求解步骤

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

使用换元积分法

= \int \frac{cos ( n u ) cos ( \frac{u}{2} )}{2} d u

提出常数:

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

= \frac{1}{2} \cdot \int cos (n u) cos (\frac{u}{2}) d u

使用分布积分法

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

\int - \frac{1}{2 n} sin (n u) sin (\frac{u}{2}) d u = - \frac{1}{4 n} (- \frac{2}{2 n + 1} sin (n u + \frac{u}{2}) + \frac{2}{2 n - 1} sin (n u - \frac{u}{2}))

= \frac{1}{2} (\frac{1}{n} sin (n u) cos (\frac{u}{2}) - (- \frac{1}{4 n} (- \frac{2}{2 n + 1} sin (n u + \frac{u}{2}) + \frac{2}{2 n - 1} sin (n u - \frac{u}{2}))))

u = 2 x

代回

= \frac{1}{2} (\frac{1}{n} sin (n \cdot 2 x) cos (\frac{2 x}{2}) - (- \frac{1}{4 n} (- \frac{2}{2 n + 1} sin (n \cdot 2 x + \frac{2 x}{2}) + \frac{2}{2 n - 1} sin (n \cdot 2 x - \frac{2 x}{2}))))

化简

\frac{1}{2} (\frac{1}{n} sin (n \cdot 2 x) cos (\frac{2 x}{2}) - (- \frac{1}{4 n} (- \frac{2}{2 n + 1} sin (n \cdot 2 x + \frac{2 x}{2}) + \frac{2}{2 n - 1} sin (n \cdot 2 x - \frac{2 x}{2})))) : \frac{1}{2} (\frac{1}{n} sin (2 n x) cos (x) + \frac{1}{4 n} (- \frac{2}{2 n + 1} sin (2 n x + x) + \frac{2}{2 n - 1} sin (2 n x - x)))

= \frac{1}{2} (\frac{1}{n} sin (2 n x) cos (x) + \frac{1}{4 n} (- \frac{2}{2 n + 1} sin (2 n x + x) + \frac{2}{2 n - 1} sin (2 n x - x)))

解答补常数

= \frac{1}{2} (\frac{1}{n} sin (2 n x) cos (x) + \frac{1}{4 n} (- \frac{2}{2 n + 1} sin (2 n x + x) + \frac{2}{2 n - 1} sin (2 n x - x))) + C

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