zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的微积分 >

integral of 1/((sin(x))^5(cos(x))^5)

解答

\int \frac{1}{( sin ( x ) ) ^{5} ( cos ( x ) ) ^{5}} d x

解答

6 ln ∣ sin (x) ∣ - \frac{3}{2} csc^{2} (x) - \frac{1}{4} csc^{4} (x) - 3 ln ∣ sin (x) + 1 ∣ + \frac{13}{16 ( sin ( x ) + 1 )} + \frac{1}{16 ( sin ( x ) + 1 ) ^{2}} - 3 ln ∣ sin (x) - 1 ∣ - \frac{13}{16 ( sin ( x ) - 1 )} + \frac{1}{16 ( sin ( x ) - 1 ) ^{2}} + C

求解步骤

\int \frac{1}{( sin ( x ) ) ^{5} ( cos ( x ) ) ^{5}} d x

化简

= \int \frac{1}{sin ^{5} ( x ) cos ^{5} ( x )} d x

使用换元积分法

= \int \frac{1}{u ^{5} ( 1 - u ^{2} ) ^{3}} d u

将

\frac{1}{u ^{5} ( 1 - u ^{2} ) ^{3}}

用部份分式展开:

\frac{6}{u} + \frac{3}{u ^{3}} + \frac{1}{u ^{5}} - \frac{3}{u + 1} - \frac{13}{16 ( u + 1 ) ^{2}} - \frac{1}{8 ( u + 1 ) ^{3}} - \frac{3}{u - 1} + \frac{13}{16 ( u - 1 ) ^{2}} - \frac{1}{8 ( u - 1 ) ^{3}}

= \int \frac{6}{u} + \frac{3}{u ^{3}} + \frac{1}{u ^{5}} - \frac{3}{u + 1} - \frac{13}{16 ( u + 1 ) ^{2}} - \frac{1}{8 ( u + 1 ) ^{3}} - \frac{3}{u - 1} + \frac{13}{16 ( u - 1 ) ^{2}} - \frac{1}{8 ( u - 1 ) ^{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

= \int \frac{6}{u} d u + \int \frac{3}{u ^{3}} d u + \int \frac{1}{u ^{5}} d u - \int \frac{3}{u + 1} d u - \int \frac{13}{16 ( u + 1 ) ^{2}} d u - \int \frac{1}{8 ( u + 1 ) ^{3}} d u - \int \frac{3}{u - 1} d u + \int \frac{13}{16 ( u - 1 ) ^{2}} d u - \int \frac{1}{8 ( u - 1 ) ^{3}} d u

\int \frac{6}{u} d u = 6 ln ∣ u ∣

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

\int \frac{1}{u ^{5}} d u = - \frac{1}{4 u ^{4}}

\int \frac{3}{u + 1} d u = 3 ln ∣ u + 1 ∣

\int \frac{13}{16 ( u + 1 ) ^{2}} d u = - \frac{13}{16 ( u + 1 )}

\int \frac{1}{8 ( u + 1 ) ^{3}} d u = - \frac{1}{16 ( u + 1 ) ^{2}}

\int \frac{3}{u - 1} d u = 3 ln ∣ u - 1 ∣

\int \frac{13}{16 ( u - 1 ) ^{2}} d u = - \frac{13}{16 ( u - 1 )}

\int \frac{1}{8 ( u - 1 ) ^{3}} d u = - \frac{1}{16 ( u - 1 ) ^{2}}

= 6 ln ∣ u ∣ - \frac{3}{2 u ^{2}} - \frac{1}{4 u ^{4}} - 3 ln ∣ u + 1 ∣ - (- \frac{13}{16 ( u + 1 )}) - (- \frac{1}{16 ( u + 1 ) ^{2}}) - 3 ln ∣ u - 1 ∣ - \frac{13}{16 ( u - 1 )} - (- \frac{1}{16 ( u - 1 ) ^{2}})

u = sin (x)

代回

= 6 ln ∣ sin (x) ∣ - \frac{3}{2 sin ^{2} ( x )} - \frac{1}{4 sin ^{4} ( x )} - 3 ln ∣ sin (x) + 1 ∣ - (- \frac{13}{16 ( sin ( x ) + 1 )}) - (- \frac{1}{16 ( sin ( x ) + 1 ) ^{2}}) - 3 ln ∣ sin (x) - 1 ∣ - \frac{13}{16 ( sin ( x ) - 1 )} - (- \frac{1}{16 ( sin ( x ) - 1 ) ^{2}})

化简

6 ln ∣ sin (x) ∣ - \frac{3}{2 sin ^{2} ( x )} - \frac{1}{4 sin ^{4} ( x )} - 3 ln ∣ sin (x) + 1 ∣ - (- \frac{13}{16 ( sin ( x ) + 1 )}) - (- \frac{1}{16 ( sin ( x ) + 1 ) ^{2}}) - 3 ln ∣ sin (x) - 1 ∣ - \frac{13}{16 ( sin ( x ) - 1 )} - (- \frac{1}{16 ( sin ( x ) - 1 ) ^{2}}) : 6 ln ∣ sin (x) ∣ - \frac{3}{2} csc^{2} (x) - \frac{1}{4} csc^{4} (x) - 3 ln ∣ sin (x) + 1 ∣ + \frac{13}{16 ( sin ( x ) + 1 )} + \frac{1}{16 ( sin ( x ) + 1 ) ^{2}} - 3 ln ∣ sin (x) - 1 ∣ - \frac{13}{16 ( sin ( x ) - 1 )} + \frac{1}{16 ( sin ( x ) - 1 ) ^{2}}

= 6 ln ∣ sin (x) ∣ - \frac{3}{2} csc^{2} (x) - \frac{1}{4} csc^{4} (x) - 3 ln ∣ sin (x) + 1 ∣ + \frac{13}{16 ( sin ( x ) + 1 )} + \frac{1}{16 ( sin ( x ) + 1 ) ^{2}} - 3 ln ∣ sin (x) - 1 ∣ - \frac{13}{16 ( sin ( x ) - 1 )} + \frac{1}{16 ( sin ( x ) - 1 ) ^{2}}

解答补常数

= 6 ln ∣ sin (x) ∣ - \frac{3}{2} csc^{2} (x) - \frac{1}{4} csc^{4} (x) - 3 ln ∣ sin (x) + 1 ∣ + \frac{13}{16 ( sin ( x ) + 1 )} + \frac{1}{16 ( sin ( x ) + 1 ) ^{2}} - 3 ln ∣ sin (x) - 1 ∣ - \frac{13}{16 ( sin ( x ) - 1 )} + \frac{1}{16 ( sin ( x ) - 1 ) ^{2}} + C

作图

流行的例子

inverselaplace 1/(19s^2+5s+19990)

in v erse l a pl a ce \frac{1}{19 s ^{2} + 5 s + 19990}

y^{''}+6y^'+45y=0

y^{^{''}} + 6 y^{^{'}} + 45 y = 0

derivative y=(x+9)(3sqrt(x)+4)

d er i v a t i v e y = (x + 9) (3 x + 4)

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

\int (3 x - 2 3 x) d x

integral of (x-3y^2)

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