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人気のある微分積分 >

integral of cos^6(4θ)

解

\int cos^{6} (4 θ) d θ

解

\frac{1}{4} (\frac{1}{6} cos^{5} (4 θ) sin (4 θ) + \frac{5}{6} (\frac{1}{4} cos^{3} (4 θ) sin (4 θ) + \frac{3}{8} (4 θ + \frac{1}{2} sin (8 θ)))) + C

解答ステップ

\int cos^{6} (4 θ) d θ

u置換積分法を適用する

= \int cos^{6} (u) \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 cos^{6} (u) d u

積分換算を適用する

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

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

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

代用を戻す

u = 4 θ

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

簡素化

\frac{1}{4} (\frac{sin ( 4 θ ) cos ^{5} ( 4 θ )}{6} + \frac{5}{6} (\frac{1}{4} cos^{3} (4 θ) sin (4 θ) + \frac{3}{8} (4 θ + \frac{1}{2} sin (2 \cdot 4 θ)))) : \frac{1}{4} (\frac{1}{6} cos^{5} (4 θ) sin (4 θ) + \frac{5}{6} (\frac{1}{4} cos^{3} (4 θ) sin (4 θ) + \frac{3}{8} (4 θ + \frac{1}{2} sin (8 θ))))

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

定数を解答に追加する

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

グラフ

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