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受欢迎的微积分 >

integral of sqrt(1+sin(θ))

解答

\int 1 + sin (θ) d θ

解答

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

求解步骤

\int 1 + sin (θ) d θ

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{θ}{2})

代回

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

化简

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

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

解答补常数

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

作图

流行的例子

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\int \frac{w ^{2} + 4 w - 1}{w ^{3} - w} d w

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\int \frac{e ^{\frac{1}{x + 1}}}{( x + 1 ) ^{2}} d x

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\int \frac{ln ^{3} ( 1 - x )}{1 - x} d x

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\int (x^{5} + 4 x^{2}) (x^{3} + 1)^{12} d x

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\int \frac{9 x ^{2}}{( 16 + x ^{2} ) ^{2}} d x