What is the antiderivative of #1/sinx#?
2 Answers
It is
Explanation:
# = (csc^2 x + csc x cot x)/(cscx+cotx)#
The numerator is the opposite (the 'negative') of the derivative of the denomoinator.
So the antiderivative is minus the natural logarithm of the denominator.
(If you've learned the technique of substitution, we can use
You can verify this answer by differentiating.
A different approach to it
Substitute
#1/(1-u^2)=1/((u-1)(u+1))=A/(u-1)+B/(u+1)# #=#
We need
Therefore,