0
votes
1answer
42 views

Logarithms with trigonometric inequality

My class is going to have an exam tomorrow, but we can't figure out how to solve such equations. $$\log_{\ \large tg(x)} \sqrt{\sin(x)^2 - 5/12} < 1 $$ We tried to transform $1$ to $\log_{\ ...
0
votes
2answers
60 views

Are the sums equal to each other?

They are $2$ different results for the integral $$\int xe^{2x}\sin\left(\frac x3\right)\,dx$$ $\displaystyle\frac{-3}{1369}e^{2x}\left(3(35-74x)\sin\left(\frac x3\right)+(37x-36)\cos\left(\frac ...
0
votes
2answers
87 views

How do I solve this integral?

As stated the title, I get to a point which I can't do anything, and I'm sure I've made a mistake some where, here is my full working out: $$ \int e^{ix}\cos(x)dx \\ u = e^{ix} \text{ | } u'= ie^{ie} ...
2
votes
1answer
76 views

Can someone please simplify this, please.

After solving my previous question, click here for question page, I tried to go up a notch and complicate the question just a bit further, turns out $\int e^x\sin(x)\cos(x)dx$ is much more different ...
1
vote
3answers
101 views

Why $\int{\log^2(2\sin(\pi x))}dx\neq\frac{\log^3(2\sin(\pi x))\sin(\pi x)}{3\pi \cos(\pi x)}$?

Apparently I completely forgot the basics of calculus after 13 years of not studying it. Why won't: $$\frac{d \log^2(2\sin(\pi x))}{dx} = \frac{\log^3(2\sin(\pi x))\sin(\pi x)}{3\pi \cos(\pi x)}$$ ...
0
votes
3answers
197 views

Why is $-\ln(\cos(x))$ equal to $\ln(\sec(x))$?

Why does the value $-\ln(|\cos(x)|)$ become $\ln(|\sec(x)|)?$ I was doing an integral and I got my final answer as that, but I don't understand how you can just send the negative sign inside and make ...
16
votes
2answers
665 views

Integral $\int_0^{\pi/2}\frac{x}{\sin x}\log^2\left(\frac{1+\cos x-\sin x}{1+\cos x+\sin x}\right)dx$

Please help me to evaluate this integral: $$\large\int_0^{\pi/2}\frac{x}{\sin x}\log^2\left(\frac{1+\cos x-\sin x}{1+\cos x+\sin x}\right)dx$$
9
votes
1answer
132 views

An integral involving the inverse of $f(x)=\log x-\log\cos x+x\tan x$

Let the function $f:\left(0,\,\displaystyle\frac\pi2\right)\to\mathbb{R}$ be defined as $$f(x)=\log x-\log\cos x+x\tan x.$$ Let its inverse be denoted as ...
7
votes
1answer
197 views

Integral $\int_0^\infty\frac{dx}{\frac{x^4-1}{x\cos(\pi\ln x)+1}+2x^2+2}$

I need your help with this integral: $$\int_0^\infty\frac{dx}{\frac{x^4-1}{x\cos(\pi\ln x)+1}+2\,x^2+2}.$$ I wasn't able to evaluate it in a closed form, although an approximate numerical evaluation ...
1
vote
1answer
48 views

domain of function with logarithmic terms.

what will be the domain of function given below? $$y=1+3(\log(\sin(x))+\log(\csc(x)))$$ in book it is given this is valid for the values of angles of 1st and 2nd quadrant only. why this function is ...
0
votes
1answer
227 views

Evaluation of the integral $\int \cos\omega t\ln\cos\omega t\,dt$

I am trying to evaluate an integral of the form $$ \int \cos\left(\omega t\right) \ln \cos\left(\omega t\right) dt$$ and am unsure how to proceed. I rewrote it as: $$ \textrm{Re} \left\{\int dt ...
1
vote
0answers
77 views

simplification of a natural log of a trigonometric function

hope you are all well. I am having a bit of a mental block, I am wondering if it is possible to simplify the following expression: $$k\cos X \cdot 4\ln(\cos X)$$ where $k$ is a constant and $X$ ...
7
votes
3answers
331 views

Broken Calculator: only certain unary functions work.

I have run into a challenge on Codecademy.com that has me absolutely bewildered. I'm sure I'm just overlooking an obvious solution, but I've been scouring tables of trigonometric and logarithmic ...
0
votes
1answer
33 views

Logarithm checking my work.

$\ln e^{2x}=6$ 2x=6 x=3 Is this method correct. ln$e$ cancel into 1.
13
votes
4answers
440 views

How to calculate $I=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}\frac{\ln(\sin y)\ln(\cos y)}{\sin y\cos y}dy$?

How do I integrate this guy? I've been stuck on this for hours.. $$I=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}\frac{\ln(\sin y)\ln(\cos y)}{\sin y\cos y}dy$$
4
votes
2answers
238 views

Natural logarithm, equivalent function

I'm developing a software with a tool unable to "recognize" the ln(), so is there a way to get the equivalen to ln() using someones of functions below? • sin1(a) • cos1(a) • tan1(a) • ...
5
votes
3answers
496 views

Verifing $\int_0^{\pi}x\ln(\sin x)\,dx=-\ln(2){\pi}^2/2$

I used all I know to show that $$\int_0^\pi x\ln(\sin x)dx=-\ln(2) \pi^2/2$$ This is my homework but don't know where to start. I appreciate your help.
0
votes
1answer
305 views

Why does $\ln|\cot x|=\ln|\cos x|-\ln|\sin x|$ hold?

I am learning trigonometry. I can solve simple trigonometric equations. But its integration with log always confuses me. I am thinking about this sums since last two hours but can't find the ...
1
vote
4answers
164 views

How can I calculate $\lim_{x \to 0} \log(\cos(x))/\log(\cos(3x))$ without l'Hopital?

How can I calculate the following limit without using, as Wolfram Alpha does, without using l'Hôpital? $$ \lim_{x\to 0}\frac{\log\cos x}{\log\cos 3x} $$
5
votes
1answer
154 views

Why does $\cos (\pi\cos (\pi \cos (\log (20+\pi)))) \approx -1$

I read on Wikipedia that $$\cos (\pi\cos (\pi \cos (\log (20+\pi)))) \approx -1$$ to a high degree of accuracy. Why is this true? Is this pure coincidence or is there some mathematical ...
15
votes
3answers
2k views

How does e, or the exponential function, relate to rotation?

$e^{i \pi} = -1$. I get why this works from a sum-of-series perspective and from an integration perspective, as in I can evaluate the integrals and find this result. However, I don't understand it ...
2
votes
2answers
378 views

Baffling identity: $\prod \log_{10} \tan = \sum \log_{10} \tan$

I am quite a baffled now, I am not getting by how it can be written that : \begin{align*} &\log_{10} \tan 40^\circ \cdot \log_{10} \tan 41^\circ \cdot \log_{10} \tan 42^\circ \cdot\log_{10} \tan ...
8
votes
5answers
1k views

Understanding imaginary exponents

Greetings! I am trying to understand what it means to have an imaginary number in an exponent. What does $x^{i}$ where $x$ is real mean? I've read a few pages on this issue, and they all seem to ...