You can combine the Maclaurin series of the logarithm and cosine here. If you look at the Maclaurin series for $\cos(x)$, it starts with $1$.

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Apr 25, 2017 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for …

As pointed out in the comments, the integral is non-elementary i.e. the integrand doesn't possess a primitive

I want to write down $\ln(\cos(x))$ Maclaurin polynomial of degree 6. I'm having trouble understanding what I need to do, let alone explain why it's true rigorously. The known …

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the derivative of $$\ln(\cos(x))$$ is given by $$-\frac{\sin(x)}{\cos(x)}$$ after the chain rule and the derivative of $\ln(x)$ is equal to $$\frac{1}{x}$$ Share Cite

Nov 19, 2015 · I know now that the right answer is $\ln(\sec x) + C$, but the answer I put was $\ln|-\cos x| + C$ and I was wondering if that answer would also work. My logic is that the …

#"differentiate using the "color(blue)"chain rule"# #"Given "y=f(g(x))" then"# #dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"#

Feb 25, 2015 · The answer is: #ln(sqrt2+1)#. The lenght of a function written in cartesian coordinates is: #L=int_a^bsqrt(1+[f'(x)]^2)dx#.