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人気のある微分積分 >

integral of 2xsin^4(x^2-1)

解

\int 2 x sin^{4} (x^{2} - 1) d x

解

- \frac{1}{4} sin^{3} (x^{2} - 1) cos (x^{2} - 1) + \frac{3}{8} (x^{2} - 1 - \frac{1}{2} sin (2 (x^{2} - 1))) + C

解答ステップ

\int 2 x sin^{4} (x^{2} - 1) d x

定数を除く:

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

= 2 \cdot \int x sin^{4} (x^{2} - 1) d x

u置換積分法を適用する

= 2 \cdot \int \frac{sin ^{4} ( u )}{2} d u

定数を除く:

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

= 2 \cdot \frac{1}{2} \cdot \int sin^{4} (u) d u

積分換算を適用する

= 2 \cdot \frac{1}{2} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \int sin^{2} (u) d u)

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

= 2 \cdot \frac{1}{2} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u)))

代用を戻す

u = x^{2} - 1

= 2 \cdot \frac{1}{2} (- \frac{cos ( x ^{2} - 1 ) sin ^{3} ( x ^{2} - 1 )}{4} + \frac{3}{4} \cdot \frac{1}{2} (x^{2} - 1 - \frac{1}{2} sin (2 (x^{2} - 1))))

簡素化

2 \cdot \frac{1}{2} (- \frac{cos ( x ^{2} - 1 ) sin ^{3} ( x ^{2} - 1 )}{4} + \frac{3}{4} \cdot \frac{1}{2} (x^{2} - 1 - \frac{1}{2} sin (2 (x^{2} - 1)))) : - \frac{1}{4} sin^{3} (x^{2} - 1) cos (x^{2} - 1) + \frac{3}{8} (x^{2} - 1 - \frac{1}{2} sin (2 (x^{2} - 1)))

= - \frac{1}{4} sin^{3} (x^{2} - 1) cos (x^{2} - 1) + \frac{3}{8} (x^{2} - 1 - \frac{1}{2} sin (2 (x^{2} - 1)))

定数を解答に追加する

= - \frac{1}{4} sin^{3} (x^{2} - 1) cos (x^{2} - 1) + \frac{3}{8} (x^{2} - 1 - \frac{1}{2} sin (2 (x^{2} - 1))) + C

グラフ

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