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Calculus Examples
arcsin(x)ydx+(1-ey)dy=0arcsin(x)ydx+(1−ey)dy=0
Step 1
Subtract arcsin(x)ydx from both sides of the equation.
(1-ey)dy=-arcsin(x)ydx
Step 2
Multiply both sides by y.
y(1-ey)dy=y(-arcsin(x)y)dx
Step 3
Step 3.1
Apply the distributive property.
(y⋅1+y(-ey))dy=y(-arcsin(x)y)dx
Step 3.2
Multiply y by 1.
(y+y(-ey))dy=y(-arcsin(x)y)dx
Step 3.3
Rewrite using the commutative property of multiplication.
(y-yey)dy=y(-arcsin(x)y)dx
Step 3.4
Rewrite using the commutative property of multiplication.
(y-yey)dy=-yarcsin(x)ydx
Step 3.5
Cancel the common factor of y.
Step 3.5.1
Factor y out of -y.
(y-yey)dy=y⋅-1arcsin(x)ydx
Step 3.5.2
Cancel the common factor.
(y-yey)dy=y⋅-1arcsin(x)ydx
Step 3.5.3
Rewrite the expression.
(y-yey)dy=-arcsin(x)dx
(y-yey)dy=-arcsin(x)dx
(y-yey)dy=-arcsin(x)dx
Step 4
Step 4.1
Set up an integral on each side.
∫y-yeydy=∫-arcsin(x)dx
Step 4.2
Integrate the left side.
Step 4.2.1
Split the single integral into multiple integrals.
∫ydy+∫-yeydy=∫-arcsin(x)dx
Step 4.2.2
By the Power Rule, the integral of y with respect to y is 12y2.
12y2+C1+∫-yeydy=∫-arcsin(x)dx
Step 4.2.3
Since -1 is constant with respect to y, move -1 out of the integral.
12y2+C1-∫yeydy=∫-arcsin(x)dx
Step 4.2.4
Integrate by parts using the formula ∫udv=uv-∫vdu, where u=y and dv=ey.
12y2+C1-(yey-∫eydy)=∫-arcsin(x)dx
Step 4.2.5
The integral of ey with respect to y is ey.
12y2+C1-(yey-(ey+C2))=∫-arcsin(x)dx
Step 4.2.6
Simplify.
12y2-(yey-ey)+C3=∫-arcsin(x)dx
12y2-(yey-ey)+C3=∫-arcsin(x)dx
Step 4.3
Integrate the right side.
Step 4.3.1
Since -1 is constant with respect to x, move -1 out of the integral.
12y2-(yey-ey)+C3=-∫arcsin(x)dx
Step 4.3.2
Integrate by parts using the formula ∫udv=uv-∫vdu, where u=arcsin(x) and dv=1.
12y2-(yey-ey)+C3=-(arcsin(x)x-∫x1√1-x2dx)
Step 4.3.3
Combine x and 1√1-x2.
12y2-(yey-ey)+C3=-(arcsin(x)x-∫x√1-x2dx)
Step 4.3.4
Let u=1-x2. Then du=-2xdx, so -12du=xdx. Rewrite using u and du.
Step 4.3.4.1
Let u=1-x2. Find dudx.
Step 4.3.4.1.1
Differentiate 1-x2.
ddx[1-x2]
Step 4.3.4.1.2
Differentiate.
Step 4.3.4.1.2.1
By the Sum Rule, the derivative of 1-x2 with respect to x is ddx[1]+ddx[-x2].
ddx[1]+ddx[-x2]
Step 4.3.4.1.2.2
Since 1 is constant with respect to x, the derivative of 1 with respect to x is 0.
0+ddx[-x2]
0+ddx[-x2]
Step 4.3.4.1.3
Evaluate ddx[-x2].
Step 4.3.4.1.3.1
Since -1 is constant with respect to x, the derivative of -x2 with respect to x is -ddx[x2].
0-ddx[x2]
Step 4.3.4.1.3.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
0-(2x)
Step 4.3.4.1.3.3
Multiply 2 by -1.
0-2x
0-2x
Step 4.3.4.1.4
Subtract 2x from 0.
-2x
-2x
Step 4.3.4.2
Rewrite the problem using u and du.
12y2-(yey-ey)+C3=-(arcsin(x)x-∫1√u⋅1-2du)
12y2-(yey-ey)+C3=-(arcsin(x)x-∫1√u⋅1-2du)
Step 4.3.5
Simplify.
Step 4.3.5.1
Move the negative in front of the fraction.
12y2-(yey-ey)+C3=-(arcsin(x)x-∫1√u(-12)du)
Step 4.3.5.2
Multiply 1√u by 12.
12y2-(yey-ey)+C3=-(arcsin(x)x-∫-1√u⋅2du)
Step 4.3.5.3
Move 2 to the left of √u.
12y2-(yey-ey)+C3=-(arcsin(x)x-∫-12√udu)
12y2-(yey-ey)+C3=-(arcsin(x)x-∫-12√udu)
Step 4.3.6
Since -1 is constant with respect to u, move -1 out of the integral.
12y2-(yey-ey)+C3=-(arcsin(x)x--∫12√udu)
Step 4.3.7
Simplify.
Step 4.3.7.1
Multiply -1 by -1.
12y2-(yey-ey)+C3=-(arcsin(x)x+1∫12√udu)
Step 4.3.7.2
Multiply ∫12√udu by 1.
12y2-(yey-ey)+C3=-(arcsin(x)x+∫12√udu)
12y2-(yey-ey)+C3=-(arcsin(x)x+∫12√udu)
Step 4.3.8
Since 12 is constant with respect to u, move 12 out of the integral.
12y2-(yey-ey)+C3=-(arcsin(x)x+12∫1√udu)
Step 4.3.9
Apply basic rules of exponents.
Step 4.3.9.1
Use n√ax=axn to rewrite √u as u12.
12y2-(yey-ey)+C3=-(arcsin(x)x+12∫1u12du)
Step 4.3.9.2
Move u12 out of the denominator by raising it to the -1 power.
12y2-(yey-ey)+C3=-(arcsin(x)x+12∫(u12)-1du)
Step 4.3.9.3
Multiply the exponents in (u12)-1.
Step 4.3.9.3.1
Apply the power rule and multiply exponents, (am)n=amn.
12y2-(yey-ey)+C3=-(arcsin(x)x+12∫u12⋅-1du)
Step 4.3.9.3.2
Combine 12 and -1.
12y2-(yey-ey)+C3=-(arcsin(x)x+12∫u-12du)
Step 4.3.9.3.3
Move the negative in front of the fraction.
12y2-(yey-ey)+C3=-(arcsin(x)x+12∫u-12du)
12y2-(yey-ey)+C3=-(arcsin(x)x+12∫u-12du)
12y2-(yey-ey)+C3=-(arcsin(x)x+12∫u-12du)
Step 4.3.10
By the Power Rule, the integral of u-12 with respect to u is 2u12.
12y2-(yey-ey)+C3=-(arcsin(x)x+12(2u12+C4))
Step 4.3.11
Rewrite -(arcsin(x)x+12(2u12+C4)) as -(arcsin(x)x+u12)+C4.
12y2-(yey-ey)+C3=-(arcsin(x)x+u12)+C4
Step 4.3.12
Replace all occurrences of u with 1-x2.
12y2-(yey-ey)+C3=-(arcsin(x)x+(1-x2)12)+C4
Step 4.3.13
Simplify.
Step 4.3.13.1
Apply the distributive property.
12y2-(yey-ey)+C3=-(arcsin(x)x)-(1-x2)12+C4
Step 4.3.13.2
Reorder factors in -arcsin(x)x-(1-x2)12.
12y2-(yey-ey)+C3=-xarcsin(x)-(1-x2)12+C4
12y2-(yey-ey)+C3=-xarcsin(x)-(1-x2)12+C4
12y2-(yey-ey)+C3=-xarcsin(x)-(1-x2)12+C4
Step 4.4
Group the constant of integration on the right side as K.
12y2-(yey-ey)=-xarcsin(x)-(1-x2)12+K
12y2-(yey-ey)=-xarcsin(x)-(1-x2)12+K