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I'd like to ask you a question about calculus. How does SiNx integrate to the sixth power
I1=∫(sinx)^6 dx=-∫(sinx)^5 dcosx = -cosx (sinx)^5 + ∫cosxd(sinx)^5 =-cosx (sinx)^5 + 5∫cosx(sinx)^4 cosxdx =-cosx (sinx)^5 + 5∫(sinx)^4 - (sinx)^6dx6I1 =-cosx (sinx)^5 + 5∫(sinx)^4dx ----1I2 = ...
The derivative of y = y (x) determined by SiNx to the power of y = cosy to the power of X is dy / DX
Take the natural logarithm on both sides at the same time to obtain:
y*ln(sinx)=x*ln(cosy)
When both sides take the derivative of X at the same time, we get:
y'*ln(sinx)+y*cotx=ln(cosy)-y'x*tany
Solution y ':
dy/dx=y'=[ln(cosy)-y*cotx]/[ln(sinx)+x*tany]
Find the general solution of the differential equation (y + 1) square multiplication dy / DX plus the third power of x = 0
(y+1)^2 *dy/dx+x^3=0
(y+1)^2 *dy/dx=-x^3
(y+1)^2 *dy=-x^3dx
1/3*(y+1)^3=-1/4*x^4+C
(y+1)^3=-3/4*x^4+C
Solution of differential equation: XY power of dy / DX + Y / x = e won't play that power. Thank you! Just give me a hint! Is it a homogeneous equation! I think so, but I won't change!
This is not a homogeneous equation. We need to use the method of variable substitution
The left becomes (XY '+ y) / x = D (XY) / DX × 1 / x, so take XY as a new dependent variable and let u = XY, then the original equation is changed into: Du / DX × 1/x＝e^u.
Separate variables: e ^ (- U) Du = xdx
Integral on both sides: - e ^ (- U) = 1 / 2 × x^2＋C
Substituting u = XY, the general solution of the original equation is 1 / 2 × x^2＋C＋e^(-xy)＝0
Let the function y = f (x) have f '(X.), then when Δˇ X → 0f (x) at x = x ˇ The differential dy at O is A and equivalent infinitesimal B are infinitesimal with the same price, but not equivalent infinitesimal C is infinitesimal with high price and D is infinitesimal with low price
f＇(x)=dy/dx
dy=f＇(x)dx
Then when Δˇ X → 0f (x) at x = x ˇ Differential at o
dy=f＇(xo)dx
Find the differential dy of the following function: y = LNX / SiNx
y'=[(lnx)'sinx-lnx*(sinx)']/(sinx)^2
=(sinx*1/x-lnx*cosx)/(sinx)^2
So dy = (SiNx * 1 / x-lnx * cosx) / (SiNx) ^ 2DX
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