- #1
kristo
- 13
- 0
Homework Statement
Find the antiderivative of [tex]\frac{1}{\sin\alpha}[/tex]
Homework Equations
The Attempt at a Solution
Ugh, maybe it's just a temporary brainfreeze, but I have no idea how to go about it.
kristo said:Homework Statement
Find the antiderivative of [tex]\frac{1}{\sin\alpha}[/tex]
Homework Equations
The Attempt at a Solution
Ugh, maybe it's just a temporary brainfreeze, but I have no idea how to go about it.
The antiderivative of sin(-1)x is cos(x) + C, where C is a constant.
To find the antiderivative of sin(-1)x, you can use the power rule for integration, which states that the antiderivative of x^n is (x^(n+1))/(n+1). In this case, n = -1, so the antiderivative is (x^-1)/-1 = -1/x. However, since we are dealing with sin(-1)x, we need to use the chain rule. This results in the antiderivative of sin(-1)x being cos(x) + C.
Step 1: Rewrite sin(-1)x as 1/sin(x) using the reciprocal identity for sine.Step 2: Use the power rule for integration, which gives us the antiderivative of 1/sin(x) as -1/x.Step 3: Use the chain rule by substituting -1/x back into the original expression sin(-1)x, resulting in cos(x) + C.
Yes, there is a shortcut for finding the antiderivative of sin(-1)x. Since the derivative of cos(x) is -sin(x), it follows that the antiderivative of -sin(x) is cos(x). Therefore, the antiderivative of sin(-1)x is cos(x) + C.
The antiderivative of sin(-1)x is the function that, when differentiated, gives us sin(-1)x. The integral of sin(-1)x is the area under the curve of sin(-1)x between two points. In other words, the antiderivative is the reverse process of differentiation, while the integral is the reverse process of finding the area under a curve using integration.