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Calculus Examples
dydx=cot(x) , y(-3)=7
Step 1
Rewrite the equation.
dy=cot(x)dx
Step 2
Step 2.1
Set up an integral on each side.
∫dy=∫cot(x)dx
Step 2.2
Apply the constant rule.
y+C1=∫cot(x)dx
Step 2.3
The integral of cot(x) with respect to x is ln(|sin(x)|).
y+C1=ln(|sin(x)|)+C2
Step 2.4
Group the constant of integration on the right side as C.
y=ln(|sin(x)|)+C
y=ln(|sin(x)|)+C
Step 3
Use the initial condition to find the value of C by substituting -3 for x and 7 for y in y=ln(|sin(x)|)+C.
7=ln(|sin(-3)|)+C
Step 4
Step 4.1
Rewrite the equation as ln(|sin(-3)|)+C=7.
ln(|sin(-3)|)+C=7
Step 4.2
Simplify the left side.
Step 4.2.1
Simplify each term.
Step 4.2.1.1
Evaluate sin(-3).
ln(|-0.05233595|)+C=7
Step 4.2.1.2
The absolute value is the distance between a number and zero. The distance between -0.05233595 and 0 is 0.05233595.
ln(0.05233595)+C=7
ln(0.05233595)+C=7
ln(0.05233595)+C=7
Step 4.3
Subtract ln(0.05233595) from both sides of the equation.
C=7-ln(0.05233595)
C=7-ln(0.05233595)
Step 5
Step 5.1
Substitute 7-ln(0.05233595) for C.
y=ln(|sin(x)|)+7-ln(0.05233595)
Step 5.2
Use the quotient property of logarithms, logb(x)-logb(y)=logb(xy).
y=ln(|sin(x)|0.05233595)+7
Step 5.3
Simplify each term.
Step 5.3.1
Multiply by 1.
y=ln(1|sin(x)|0.05233595)+7
Step 5.3.2
Factor 0.05233595 out of 0.05233595.
y=ln(1|sin(x)|0.05233595(1))+7
Step 5.3.3
Separate fractions.
y=ln(10.05233595⋅|sin(x)|1)+7
Step 5.3.4
Divide 1 by 0.05233595.
y=ln(19.1073226|sin(x)|1)+7
Step 5.3.5
Divide |sin(x)| by 1.
y=ln(19.1073226|sin(x)|)+7
y=ln(19.1073226|sin(x)|)+7
y=ln(19.1073226|sin(x)|)+7