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受欢迎的微积分 >

integral of sqrt(1+sin(x))

解答

\int 1 + sin (x) d x

解答

2 (- cos (\frac{x}{2}) + sin (\frac{x}{2})) + C

求解步骤

\int 1 + sin (x) d x

使用换元积分法

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

提出常数:

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

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

乘开

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

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

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

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

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

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

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

代回

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

化简

2 - \frac{1}{1 + tan ^{2} ( \frac{x}{2} )} + \frac{tan ( \frac{x}{2} )}{1 + tan ^{2} ( \frac{x}{2} )} : 2 (- cos (\frac{x}{2}) + sin (\frac{x}{2}))

= 2 (- cos (\frac{x}{2}) + sin (\frac{x}{2}))

解答补常数

= 2 (- cos (\frac{x}{2}) + sin (\frac{x}{2})) + C

作图

流行的例子

derivative f(x)=sqrt(5+4e^{3x)}

d er i v a t i v e f (x) = 5 + 4 e^{3 x}

y^{''}+16y^'+64y=0,y(0)=-3,y^'(0)=28

y^{^{''}} + 16 y^{^{'}} + 64 y = 0, y (0) = - 3, y^{^{'}} (0) = 28

integral from 0 to 8 of sqrt(1+81/4 x)

\int_{0}^{8} 1 + \frac{81}{4} x d x

derivative 4x^2-6x+2

d er i v a t i v e 4 x^{2} - 6 x + 2

derivative of 5sin(x(4/(x^3)+3x))

\frac{d}{d x} (5 sin (x) (\frac{4}{x ^{3}} + 3 x))