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人気のある微分積分 >

integral of (t^2cos(2t^3))/(csc(10t^3))

解

\int \frac{t ^{2} cos ( 2 t ^{3} )}{csc ( 10 t ^{3} )} d t

解

\frac{1}{6} (- \frac{1}{12} cos (12 t^{3}) - \frac{1}{8} cos (8 t^{3})) + C

解答ステップ

\int \frac{t ^{2} cos ( 2 t ^{3} )}{csc ( 10 t ^{3} )} d t

u置換積分法を適用する

= \int \frac{cos ( 2 u )}{3 csc ( 10 u )} d u

定数を除く:

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

= \frac{1}{3} \cdot \int \frac{cos ( 2 u )}{csc ( 10 u )} d u

基本的な三角関数の公式を使用する:

\frac{1}{csc ( x )} = sin (x)

= \frac{1}{3} \cdot \int cos (2 u) sin (10 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}{3} \cdot \int \frac{sin ( 10 u + 2 u ) + sin ( 10 u - 2 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}{3} \cdot \frac{1}{2} \cdot \int sin (10 u + 2 u) + sin (10 u - 2 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}{3} \cdot \frac{1}{2} (\int sin (10 u + 2 u) d u + \int sin (10 u - 2 u) d u)

\int sin (10 u + 2 u) d u = - \frac{1}{12} cos (12 u)

\int sin (10 u - 2 u) d u = - \frac{1}{8} cos (8 u)

= \frac{1}{3} \cdot \frac{1}{2} (- \frac{1}{12} cos (12 u) - \frac{1}{8} cos (8 u))

代用を戻す

u = t^{3}

= \frac{1}{3} \cdot \frac{1}{2} (- \frac{1}{12} cos (12 t^{3}) - \frac{1}{8} cos (8 t^{3}))

簡素化

\frac{1}{3} \cdot \frac{1}{2} (- \frac{1}{12} cos (12 t^{3}) - \frac{1}{8} cos (8 t^{3})) : \frac{1}{6} (- \frac{1}{12} cos (12 t^{3}) - \frac{1}{8} cos (8 t^{3}))

= \frac{1}{6} (- \frac{1}{12} cos (12 t^{3}) - \frac{1}{8} cos (8 t^{3}))

定数を解答に追加する

= \frac{1}{6} (- \frac{1}{12} cos (12 t^{3}) - \frac{1}{8} cos (8 t^{3})) + C

グラフ

人気の例

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x \to \infty lim (x^{2} - 1)

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\int x (x + 1) d x

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\frac{\partial}{\partial y} (ln (x))

sum from n=1 to infinity of 4^n

n = 1 \sum \infty 4^{n}

tangent y=4x^2,(2,16)

t an g e n t y = 4 x^{2}, (2, 16)