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人気のある微分積分 >

integral of sin^4(5x)

解

\int sin^{4} (5 x) d x

解

\frac{1}{5} (- \frac{1}{4} sin^{3} (5 x) cos (5 x) + \frac{3}{8} (5 x - \frac{1}{2} sin (10 x))) + C

解答ステップ

\int sin^{4} (5 x) d x

u置換積分法を適用する

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

積分換算を適用する

= \frac{1}{5} (- \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))

= \frac{1}{5} (- \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 = 5 x

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

簡素化

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

= \frac{1}{5} (- \frac{1}{4} sin^{3} (5 x) cos (5 x) + \frac{3}{8} (5 x - \frac{1}{2} sin (10 x)))

定数を解答に追加する

= \frac{1}{5} (- \frac{1}{4} sin^{3} (5 x) cos (5 x) + \frac{3}{8} (5 x - \frac{1}{2} sin (10 x))) + C

グラフ

人気の例

limit as x approaches 1+of (-1)/(x^2-1)

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

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

\int \frac{x}{( x ^{2} + 4 ) ^{2}} d x

derivative-(11)/(s^4)

d er i v a t i v e - \frac{11}{s ^{4}}

sum from n=1 to infinity of (nx^n)/(5^n)

n = 1 \sum \infty \frac{n x ^{n}}{5 ^{n}}

tangent f(x)=(x^3+5)/(x^2+1),\at x=-1

t an g e n t f (x) = \frac{x ^{3} + 5}{x ^{2} + 1}, at x = - 1