Answer: herr
Step-by-step explanation:
https://answers.yahoo.com/question/index?qid=20110127212721AAhjSZS
Answer: 4/5
Step-by-step explanation:
Answer:
yes they are equivalent
Step-by-step explanation:
when you do are going to put 1/8 = 2/16 and just imagine g = 1 :)
I hope I helped ;)
![\int \frac{dx}{x} = log(x) + c \\ \int \frac{f \prime(x)}{f(x)} = log \mid \: f(x) \mid + c \\ \int \: tan(x)dx = \int \: \frac{sin(x)}{cos(x)} dx \\ = - log|cos(x)| + c \\ = log |sec(x)| + c \\ the \: same \: rule \: goes \: for \: cot...etc.\\\int {sec}^{2}(x)dx=|tan(x)|+c\\let u=tanx\\ \frac{du}{dx}={sec}^{2}(x)\rightarrow {du}={sec}^{2}(x)dx\\ \int du=|u|+c\\ \therefore tan|x|+c](https://tex.z-dn.net/?f=%20%5Cint%20%5Cfrac%7Bdx%7D%7Bx%7D%20%20%3D%20log%28x%29%20%2B%20c%20%5C%5C%20%20%5Cint%20%5Cfrac%7Bf%20%5Cprime%28x%29%7D%7Bf%28x%29%7D%20%20%3D%20log%20%20%5Cmid%20%5C%3A%20f%28x%29%20%5Cmid%20%2B%20c%20%5C%5C%20%20%5Cint%20%5C%3A%20tan%28x%29dx%20%3D%20%20%5Cint%20%5C%3A%20%20%5Cfrac%7Bsin%28x%29%7D%7Bcos%28x%29%7D%20dx%20%5C%5C%20%20%3D%20%20-%20log%7Ccos%28x%29%7C%20%20%2B%20c%20%20%5C%5C%20%20%3D%20log%20%7Csec%28x%29%7C%20%20%2B%20c%20%5C%5C%20the%20%5C%3A%20same%20%5C%3A%20rule%20%5C%3A%20goes%20%5C%3A%20for%20%5C%3A%20cot...etc.%5C%5C%5Cint%20%7Bsec%7D%5E%7B2%7D%28x%29dx%3D%7Ctan%28x%29%7C%2Bc%5C%5Clet%20u%3Dtanx%5C%5C%20%5Cfrac%7Bdu%7D%7Bdx%7D%3D%7Bsec%7D%5E%7B2%7D%28x%29%5Crightarrow%20%7Bdu%7D%3D%7Bsec%7D%5E%7B2%7D%28x%29dx%5C%5C%20%5Cint%20du%3D%7Cu%7C%2Bc%5C%5C%20%5Ctherefore%20tan%7Cx%7C%2Bc)
Step-by-step explanation:
Am not sure what your question is? But if you are asking about a proof, then you may use Taylor series to prove these integrals...