integral of xsin^2(ax)

解答

\int x sin^{2} (a x) d x

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

求解步骤

\int x sin^{2} (a x) d x

使用换元积分法

= \int \frac{u sin ^{2} ( u )}{a ^{2}} d u

提出常数:

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

= \frac{1}{a ^{2}} \cdot \int u sin^{2} (u) d u

使用分布积分法

= \frac{1}{a ^{2}} (\frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \int \frac{1}{2} (u - \frac{1}{2} sin (2 u)) d u)

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

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

u = a x

代回

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

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

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

解答补常数

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