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人気のある微分積分 >

integral of 7sin^2(t)cos^4(t)

解

\int 7 sin^{2} (t) cos^{4} (t) d t

解

7 (\frac{3}{8} t - \frac{3}{32} sin (4 t) + \frac{19}{24} cos^{3} (t) sin (t) - \frac{1}{6} cos^{5} (t) sin (t) - \frac{5}{32} (2 t + sin (2 t))) + C

解答ステップ

\int 7 sin^{2} (t) cos^{4} (t) d t

定数を除く:

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

= 7 \cdot \int sin^{2} (t) cos^{4} (t) d t

三角関数の公式を使用して書き換える

= 7 \cdot \int (1 - cos^{2} (t)) cos^{4} (t) d t

拡張

(1 - cos^{2} (t)) cos^{4} (t) : cos^{4} (t) - cos^{6} (t)

= 7 \cdot \int cos^{4} (t) - cos^{6} (t) d t

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

= 7 (\int cos^{4} (t) d t - \int cos^{6} (t) d t)

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

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

= 7 (cos^{3} (t) sin (t) + \frac{3}{8} (t - \frac{1}{4} sin (4 t)) - (\frac{sin ( t ) cos ^{5} ( t )}{6} + \frac{5}{6} (\frac{1}{4} cos^{3} (t) sin (t) + \frac{3}{8} (t + \frac{1}{2} sin (2 t)))))

簡素化

7 (cos^{3} (t) sin (t) + \frac{3}{8} (t - \frac{1}{4} sin (4 t)) - (\frac{sin ( t ) cos ^{5} ( t )}{6} + \frac{5}{6} (\frac{1}{4} cos^{3} (t) sin (t) + \frac{3}{8} (t + \frac{1}{2} sin (2 t))))) : 7 (\frac{3}{8} t - \frac{3}{32} sin (4 t) + \frac{19}{24} cos^{3} (t) sin (t) - \frac{1}{6} cos^{5} (t) sin (t) - \frac{5}{32} (2 t + sin (2 t)))

= 7 (\frac{3}{8} t - \frac{3}{32} sin (4 t) + \frac{19}{24} cos^{3} (t) sin (t) - \frac{1}{6} cos^{5} (t) sin (t) - \frac{5}{32} (2 t + sin (2 t)))

定数を解答に追加する

= 7 (\frac{3}{8} t - \frac{3}{32} sin (4 t) + \frac{19}{24} cos^{3} (t) sin (t) - \frac{1}{6} cos^{5} (t) sin (t) - \frac{5}{32} (2 t + sin (2 t))) + C

グラフ

人気の例

laplacetransform ((2te^{-3t}))/5+3t*sin(t)

l a pl a ce t r an s f or m \frac{( 2 t e ^{- 3 t} )}{5} + 3 t \cdot sin (t)

(13+x^2)y^'-2xy=0

(13 + x^{2}) y^{^{'}} - 2 x y = 0

(\partial)/(\partial x)(yx+2x+{g}(y))

\frac{\partial}{\partial x} (y x + 2 x + g (y))

y^{^{'}} = - x y

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

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