integral from 2 to 1/x of sin^4(t)

解答

\int_{2}^{\frac{1}{x}} sin^{4} (t) d t

\frac{- sin ^{3} ( \frac{1}{x} ) cos ( \frac{1}{x} ) + sin ^{3} ( 2 ) cos ( 2 ) - 3}{4} + \frac{3}{8 x} - \frac{3}{16} sin (\frac{2}{x}) + \frac{3}{16} sin (4)

求解步骤

\int_{2}^{\frac{1}{x}} sin^{4} (t) d t

使用降次积分法

= - [\frac{cos ( t ) sin ^{3} ( t )}{4}]_{2}^{\frac{1}{x}} + \frac{3}{4} \cdot \int_{2}^{\frac{1}{x}} sin^{2} (t) d t

\int_{2}^{\frac{1}{x}} sin^{2} (t) d t = \frac{1}{2 x} - 1 - \frac{1}{4} sin (\frac{2}{x}) + \frac{1}{4} sin (4)

= - [\frac{cos ( t ) sin ^{3} ( t )}{4}]_{2}^{\frac{1}{x}} + \frac{3}{4} (\frac{1}{2 x} - 1 - \frac{1}{4} sin (\frac{2}{x}) + \frac{1}{4} sin (4))

带入上下限计算后得:

\frac{cos ( \frac{1}{x} ) sin ^{3} ( \frac{1}{x} )}{4} - \frac{cos ( 2 ) sin ^{3} ( 2 )}{4}

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

化简

= \frac{- sin ^{3} ( \frac{1}{x} ) cos ( \frac{1}{x} ) + sin ^{3} ( 2 ) cos ( 2 ) - 3}{4} + \frac{3}{8 x} - \frac{3}{16} sin (\frac{2}{x}) + \frac{3}{16} sin (4)

\int_{- 1}^{1} (3 - ∣ x ∣) d x

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