How to integrate $\ln(2+\sin x)$ I wonder how to solve this. Once I saw $\int{\ln \sin}$ is an improper integral that only possible to integrate in certain interval. And frankly that is still far from where I'm now. I'm just curious and sought for it back then.
Then now, I got a worksheet that contains playful and interesting type of problems but it's doing bad job at explaining how things worked out. So, can anyone help how to solve the problem above, just from where that answer is obtained?
note: below is an official key sheet
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 A: Because you are given some choices, you can think about it this way, (though ofcourse it only works for these kinds of given choices, but it may help)
$-1\le \sin x \le 1$
$\implies 0\le \ln(2+\sin x) \le \ln 3$
$\implies 0\le \int_3^5\ln(2+\sin x) dx\le 2.197$
So $4\le f(5) \le 6.197$
Only the answer $4.555$ satisfy this condition.
A: The remark about a calculator being permitted suggests just evaluating the integral numerically using your calculator.
Here's a method that doesn't require using a calculator at all: Since $\sin$ is decreasing on the interval $[3, 5]$ and $\frac{5 \pi}{6} < 3 < \pi$ (the first inequality is equivalent to $\pi < \frac{18}{5} = 3.6$), we have on that interval that $$-1 \leq \sin x \leq \frac{1}{2}$$ and thus $$0 \leq \log (2 + \sin x) \leq \log \frac{5}{2} .$$ Thus,
$$f(5) = f(3) + \int_3^5 f'(x) \,dx = f(3) + \int_3^5 \log (2 + \sin x) \,dx$$ is bounded between $f(3) = 4$ and $f(3) + (5 - 3) \log \frac{5}{2} = 4 + 2 \log \frac{5}{2}$. Now, $\frac{5}{2} = 1 + 1 + \frac{1}{2} < e$, so $4 + 2 \log \frac{5}{2} < 4 + 2 \log e = 4 + 2 = 6$, giving
$$4 \leq f(5) < 6 .$$
A: Although the MCQ answers just need loose bounds for the integral that can be accomplished by bounding $f'$ by constants, we can easily create sharper bounds that just require a calculator for $\cos5$ and $\cos3$.
$\ln x \leq x-1\Rightarrow I=\int_{3}^{5}\ln\left(2+\sin t\right)dt\leq \int_{3}^{5}\left(1+\sin t\right)dt=0.726$
$\ln x\geq 1-\frac{1}{x}\Rightarrow I\geq \int_{3}^{5}\frac{1+\sin t}{2+\sin t}dt \geq \int_{3}^{5}\frac{1+\sin t}{3}dt=0.242$.
Therefore $4.242\leq f(5) \leq 4.726$, which implies (D) is correct.
