hello
i know that deriv(cscx) = -cscxcotx
and that
deriv(cos) = -sinx
deriv(cotx) = -((cscx)^2)
my question is:
is this (statements below) correct?
antideriv(cscxcotx) = -(cscx)
antideriv(sinx) = -cosx
antideriv((cscx)^2)= -cotx
were those antiderivatives above correct?
ALSO basically my question is how i can remember where the negative sign goes and with which trig funcation it goes with; also is this correct : to go from a derivative to an antiderivative, (visually) change the derivative symbol to an antiderivative symbol, then switch the places of the trig identities on either side of the equal sign (flipflop them), and keep the negative sign where it is....? is this right? thank you!
1. THey are right.
2. How can you remember? Flash cards.
3. I don't like that rule, complicated. I just recommend memorizing the derivatives, when that is done, the inverse operation is easy. Flash cards.
Hello! Let's go through your questions one by one.
1. antideriv(cscxcotx) = -(cscx):
No, this statement is not correct. The antiderivative of csc(x)cot(x) is -csc(x), not -(cscx). When finding antiderivatives, you do not distribute the negative sign across the trigonometric function.
2. antideriv(sinx) = -cosx:
Yes, this statement is correct. The antiderivative of sin(x) is -cos(x).
3. antideriv((cscx)^2) = -cotx:
No, this statement is not correct. The antiderivative of (csc(x))^2 is -csc(x). Similar to the first statement, you do not distribute the negative sign across the trigonometric function.
To remember where the negative sign goes when taking derivatives and antiderivatives, you can use the following guidelines:
- When taking derivatives and encountering a trigonometric function, you apply the negative sign to the derivative. For example, when taking the derivative of csc(x), you get -csc(x)cot(x).
- When taking antiderivatives, you do not distribute the negative sign across the trigonometric function. Instead, you keep the negative sign in front of the resulting antiderivative. For example, the antiderivative of csc(x)cot(x) is -csc(x).
Regarding your visual method, the idea of switching the places of trigonometric identities and keeping the negative sign where it is not accurate. The process of going from the derivative to the antiderivative typically involves integrating the function, which is more complex than switching places and flipping signs.
In summary, remember that when taking derivatives, the negative sign is applied to the derivative itself. And when taking antiderivatives, the negative sign goes in front of the resulting antiderivative.