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人気のある微分積分 >

integral of sqrt(2+2sin(x))

解

\int 2 + 2 sin (x) d x

解

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

解答ステップ

\int 2 + 2 sin (x) d x

因数

2 + 2 sin (x) : 2 (sin (x) + 1)

= \int 2 (sin (x) + 1) d x

累乗根の規則を適用する:

n ab = n a n b,

, 以下を想定

a \geq 0, b \geq 0

= \int 2 sin (x) + 1 d x

定数を除く:

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

= 2 \cdot \int sin (x) + 1 d x

u置換積分法を適用する

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

定数を除く:

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

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

拡張

\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}}

総和規則を適用する:

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

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

代用を戻す

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

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

簡素化

2 \cdot 2 - \frac{1}{1 + tan ^{2} ( \frac{x}{2} )} + \frac{tan ( \frac{x}{2} )}{1 + tan ^{2} ( \frac{x}{2} )} : 22 (- cos (\frac{x}{2}) + sin (\frac{x}{2}))

= 22 (- cos (\frac{x}{2}) + sin (\frac{x}{2}))

定数を解答に追加する

= 22 (- cos (\frac{x}{2}) + sin (\frac{x}{2})) + C

グラフ

人気の例

integral of 2+1/x

\int 2 + \frac{1}{x} d x

tangent f(x)=(-8x)/(x^2+1),\at x=1

t an g e n t f (x) = \frac{- 8 x}{x ^{2} + 1}, at x = 1

(\partial)/(\partial x)(2x^2y+4)

\frac{\partial}{\partial x} (2 x^{2} y + 4)

勾配 (0,2),(1,0)

s l o p e (0, 2), (1, 0)

inverselaplace (4se^{-2s})/(3s^2+12)

in v erse l a pl a ce \frac{4 s e ^{- 2 s}}{3 s ^{2} + 12}