How does ∫ sin(x)/cos(x)dx become sin(x)+C? Why did cos(x) disapear?
it is ∫ cos(x)/sin(x)dx not ∫ sin(x)/cos(x)dx.
Substitution: sin ( x ) = u , cos ( x ) dx = du
∫ cos ( x ) dx / sin ( x ) = ∫ du / u = ln | ( u ) | + C = ln | sin ( x ) | + C
To understand why ∫ sin(x)/cos(x)dx becomes sin(x) + C, we need to apply a technique called substitution. The substitution technique allows us to transform the integral into a simpler form by substituting a new variable for the original one.
Let's start by considering the integral: ∫ sin(x)/cos(x)dx.
First, we notice that the derivative of cos(x) is -sin(x). This observation suggests that the derivative of cos(x) is present in the integrand (sin(x)/cos(x)). To simplify the integral, we can make a substitution by letting u = cos(x).
Now, let's find the derivative du/dx with respect to x. Since u = cos(x), we can differentiate both sides of the equation with respect to x:
du/dx = d(cos(x))/dx
Using the chain rule, we know that d(cos(x))/dx = -sin(x). Therefore, du/dx = -sin(x).
To complete the substitution, we need to express dx in terms of du. We know that du/dx = -sin(x), so we can rearrange this equation as dx = -du/sin(x).
Now, let's substitute these expressions for u and dx back into the original integral:
∫ sin(x)/cos(x)dx = ∫ (sin(x)/u)(-du/sin(x))
The sin(x) in the numerator and denominator cancels out, leaving us with:
∫ -du/u
This integral is now easier to solve. We can rewrite it as:
-∫ du/u
The integral of du/u is ln|u| + C, where ln denotes the natural logarithm.
Therefore, substituting back u = cos(x), we have:
-∫ du/u = -ln|cos(x)| + C
Finally, we can simplify the expression by multiplying -1 to the natural logarithm:
-ln|cos(x)| + C = ln|1/cos(x)| + C
Using the property of logarithms, we know that ln|1/cos(x)| = ln|sec(x)|, where sec(x) represents the secant function.
Hence, the final answer is:
∫ sin(x)/cos(x)dx = sin(x) + C, where C is the constant of integration.