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人気のある微分積分 >

integral of (1+cos(4x))^2

解

\int (1 + cos (4 x))^{2} d x

解

\frac{1}{16} (24 x + 8 sin (4 x) + sin (8 x)) + C

解答ステップ

\int (1 + cos (4 x))^{2} d x

u置換積分法を適用する

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

定数を除く:

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

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

拡張

(1 + cos (u))^{2} : 1 + 2 cos (u) + cos^{2} (u)

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

\int 1 d u = u

\int 2 cos (u) d u = 2 sin (u)

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

= \frac{1}{4} (u + 2 sin (u) + \frac{1}{2} (u + \frac{1}{2} sin (2 u)))

代用を戻す

u = 4 x

= \frac{1}{4} (4 x + 2 sin (4 x) + \frac{1}{2} (4 x + \frac{1}{2} sin (2 \cdot 4 x)))

簡素化

\frac{1}{4} (4 x + 2 sin (4 x) + \frac{1}{2} (4 x + \frac{1}{2} sin (2 \cdot 4 x))) : \frac{1}{16} (24 x + 8 sin (4 x) + sin (8 x))

= \frac{1}{16} (24 x + 8 sin (4 x) + sin (8 x))

定数を解答に追加する

= \frac{1}{16} (24 x + 8 sin (4 x) + sin (8 x)) + C

グラフ

人気の例

integral of (2+x^2)/x

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

integral of (\sqrt[3]{x^2}+sin(3x))

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

integral of ((e^{x/2}-e^{-x/2})^2)/4

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

integral of (x^2)/(sqrt(x^3)-1)

\int \frac{x ^{2}}{x ^{3} - 1} d x

integral of \sqrt[3]{(1-2x^2)}(-4x)

\int 3 (1 - 2 x^{2}) (- 4 x) d x