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受欢迎的微积分 >

integral of (2cos(x)+3sin(x))/(sin^3(x))

解答

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

解答

\frac{1}{4} (- tan^{2} (\frac{x}{2}) + 6 tan (\frac{x}{2}) - 6 cot (\frac{x}{2}) - cot^{2} (\frac{x}{2})) + C

求解步骤

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

使用换元积分法

= \int \frac{( - u ^{2} + 3 u + 1 ) ( u ^{2} + 1 )}{2 u ^{3}} d u

提出常数:

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

= \frac{1}{2} \cdot \int \frac{( - u ^{2} + 3 u + 1 ) ( u ^{2} + 1 )}{u ^{3}} d u

乘开

\frac{( - u ^{2} + 3 u + 1 ) ( u ^{2} + 1 )}{u ^{3}} : - u + 3 + \frac{3}{u ^{2}} + \frac{1}{u ^{3}}

= \frac{1}{2} \cdot \int - u + 3 + \frac{3}{u ^{2}} + \frac{1}{u ^{3}} d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (- \int u d u + \int 3 d u + \int \frac{3}{u ^{2}} d u + \int \frac{1}{u ^{3}} d u)

\int u d u = \frac{u ^{2}}{2}

\int 3 d u = 3 u

\int \frac{3}{u ^{2}} d u = - \frac{3}{u}

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

= \frac{1}{2} (- \frac{u ^{2}}{2} + 3 u - \frac{3}{u} - \frac{1}{2 u ^{2}})

u = tan (\frac{x}{2})

代回

= \frac{1}{2} (- \frac{tan ^{2} ( \frac{x}{2} )}{2} + 3 tan (\frac{x}{2}) - \frac{3}{tan ( \frac{x}{2} )} - \frac{1}{2 tan ^{2} ( \frac{x}{2} )})

化简

\frac{1}{2} (- \frac{tan ^{2} ( \frac{x}{2} )}{2} + 3 tan (\frac{x}{2}) - \frac{3}{tan ( \frac{x}{2} )} - \frac{1}{2 tan ^{2} ( \frac{x}{2} )}) : \frac{1}{4} (- tan^{2} (\frac{x}{2}) + 6 tan (\frac{x}{2}) - 6 cot (\frac{x}{2}) - cot^{2} (\frac{x}{2}))

= \frac{1}{4} (- tan^{2} (\frac{x}{2}) + 6 tan (\frac{x}{2}) - 6 cot (\frac{x}{2}) - cot^{2} (\frac{x}{2}))

解答补常数

= \frac{1}{4} (- tan^{2} (\frac{x}{2}) + 6 tan (\frac{x}{2}) - 6 cot (\frac{x}{2}) - cot^{2} (\frac{x}{2})) + C

作图

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