Calculate the curve integral I = ∮ L (y-e ^ x) DX + (3x + e ^ y) dy, where l is the positive direction of ellipse x ^ 2 / A ^ 2 + y ^ / b ^ 2 = 1

Calculate the curve integral I = ∮ L (y-e ^ x) DX + (3x + e ^ y) dy, where l is the positive direction of ellipse x ^ 2 / A ^ 2 + y ^ / b ^ 2 = 1

According to Green's formula
I=∮L(y-e^x)dx+(3x+e^y)dy＝∫∫(3-1)dxdy=2∫∫dxdy=2πab

Find the special solution of the first order linear differential equation dy / DX + ytanx = secx satisfying the initial condition y|x = 0 = 0 Using the formula method, the solution process after substitution is required. The final answer is y = SiNx,

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Y = (COS x) ^ x, find dy / DX

The best way is to find the logarithm:
LNY = xlncosx, calculate the derivative on both sides to obtain:
y'/y=lncosx-x(sinx/cosx)=lncosx-xtanx
So: y '= y (lncosx xtanx)
=(cos x)^x (lncosx-xtanx)

E ^ (- y ^ 2) + cos (x ^ 2) = 0 find dy / DX It is the derivation of implicit function

e^(-y^2)=-cos(x^2)
After the first derivation: e ^ (- y ^ 2) * - 2Y) * (dy / DX) = 2x * sin (x ^ 2)
dy/dx=-[x*sin(x^2)]/[y*e^(-y^2)]

Let y = y (x) be determined by the equation cos (x + y) + y = 1, and find dy / DX

Derivation on both sides: [- sin (x + y)] (1 + dy / DX) + dy / DX = 0 - sin (x + y) - [sin (x + y)] dy / DX + dy / DX = 0 dy / DX = [sin (x + y)] / [1-sin (x + y)]

Find the following indefinite integral 1. ∫ SiNx / (1 + SiNx) DX 2. ∫ (xcosx) / sin ² xdx

∫sinxdx/(1+sinx)=∫dx-∫dx/(1+sinx) 1+sinx=1+cos(π/2-x)=2cos(π/4-x/2)^2=∫dx-∫d(x/2)/cos(π/4-x/2)^2=x+tan(π/4-x/2)+C ∫xcosxdx/(sinx)^2=∫xd(-1/sinx)=x*(-1/sinx)+∫dx/sinx=-x/sinx-(1/2)ln|1+cosx...