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人気のある微分積分 >

integral of sin^6(5x)

解

\int sin^{6} (5 x) d x

解

- \frac{1}{30} sin^{5} (5 x) cos (5 x) + \frac{1}{96} (- 4 sin^{3} (5 x) cos (5 x) + 3 (10 x - sin (10 x))) + C

解答ステップ

\int sin^{6} (5 x) d x

u置換積分法を適用する

= \int sin^{6} (u) \frac{1}{5} d u

定数を除く:

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

= \frac{1}{5} \cdot \int sin^{6} (u) d u

積分換算を適用する

= \frac{1}{5} (- \frac{cos ( u ) sin ^{5} ( u )}{6} + \frac{5}{6} \cdot \int sin^{4} (u) d u)

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

= \frac{1}{5} (- \frac{cos ( u ) sin ^{5} ( u )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (u) cos (u) + \frac{3}{8} (u - \frac{1}{2} sin (2 u))))

代用を戻す

u = 5 x

= \frac{1}{5} (- \frac{cos ( 5 x ) sin ^{5} ( 5 x )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (5 x) cos (5 x) + \frac{3}{8} (5 x - \frac{1}{2} sin (2 \cdot 5 x))))

簡素化

\frac{1}{5} (- \frac{cos ( 5 x ) sin ^{5} ( 5 x )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (5 x) cos (5 x) + \frac{3}{8} (5 x - \frac{1}{2} sin (2 \cdot 5 x)))) : - \frac{1}{30} sin^{5} (5 x) cos (5 x) + \frac{1}{96} (- 4 sin^{3} (5 x) cos (5 x) + 3 (10 x - sin (10 x)))

= - \frac{1}{30} sin^{5} (5 x) cos (5 x) + \frac{1}{96} (- 4 sin^{3} (5 x) cos (5 x) + 3 (10 x - sin (10 x)))

定数を解答に追加する

= - \frac{1}{30} sin^{5} (5 x) cos (5 x) + \frac{1}{96} (- 4 sin^{3} (5 x) cos (5 x) + 3 (10 x - sin (10 x))) + C

グラフ

人気の例

integral of 1/((sin(x))^2)

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

sum from n=1 to infinity of (2/3)^{n+1}

n = 1 \sum \infty (\frac{2}{3})^{n + 1}

derivative f(x)=sin(x^2+2x+3)

d er i v a t i v e f (x) = sin (x^{2} + 2 x + 3)

面積 y=2x^2-1,[0,1]

a re a y = 2 x^{2} - 1, [0, 1]

tangent y=(x^2)/(2-x^2)

t an g e n t y = \frac{x ^{2}}{2 - x ^{2}}