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人気のある微分積分 >

integral of xsin(x^2)cos(3x^2)

解

\int x sin (x^{2}) cos (3 x^{2}) d x

解

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

解答ステップ

\int x sin (x^{2}) cos (3 x^{2}) d x

u置換積分法を適用する

= \int \frac{sin ( u ) cos ( 3 u )}{2} d u

定数を除く:

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

= \frac{1}{2} \cdot \int sin (u) cos (3 u) d u

次の恒等を使用する:

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

= \frac{1}{2} \cdot \int \frac{sin ( u + 3 u ) + sin ( u - 3 u )}{2} d u

定数を除く:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int sin (u + 3 u) + sin (u - 3 u) d u

総和規則を適用する:

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

= \frac{1}{2} \cdot \frac{1}{2} (\int sin (u + 3 u) d u + \int sin (u - 3 u) d u)

\int sin (u + 3 u) d u = - \frac{1}{4} cos (4 u)

\int sin (u - 3 u) d u = \frac{1}{2} cos (2 u)

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

代用を戻す

u = x^{2}

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

integral of cos(pi)

\int cos (π) d x

integral of 14tan^{13}(x)sec^2(x)

\int 14 tan^{13} (x) sec^{2} (x) d x

integral of-6sin(x)e^x

\int - 6 sin (x) e^{x} d x

integral of cos(2t)(e^{6t})

\int cos (2 t) (e^{6 t}) d t

integral of (5x^4)/(sqrt(x^5+2))

\int \frac{5 x ^{4}}{x ^{5} + 2} d x