By making the change of variable u=\pi -x you get that I=\int_0^\pi x \ln (\sin x)\:dx=\int_0^\pi (\pi-u) \ln (\sin u)\:du=\pi\int_0^\pi \ln (\sin u)\:du-I giving I=\frac {\pi}2\int_0^\pi \ln (\sin u)\:du=\pi\int_0^ {\pi/2} \ln (\sin u)\:du. ...

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$$ \int_{0}^{\pi/2}x\log\sin x\,dx = \int_{0}^{1}\frac{\arcsin(u)\log u}{\sqrt{1-u^2}}\,du\tag{1}$$ and by recalling $$ \arcsin(u) = \sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n(2n+1)}u^{2n} \tag{2} $$ $$ \int_{0}^{1}\frac{u^{2n}\log(u)}{\sqrt{1-u^2}}\,du = \frac{\pi\binom{2n}{n}}{4^{n+1}}\left(H_{n-1/2}-H_n\right)\tag{3} $$ (where $(3)$ follows by ...

For example, ∫sin (sin x)dx and ∫cos (cos x)dx can both be evaluated in series form. You can calculate ∫π 0log(sinx)dx = − πlog2 and integrating up to π / 2 would give half of this.

2016年4月20日 · 若将一点扩展成函数f (x)在其 定义域 包含的某 开区间 I内每一个点，那么函数f (x)在开区间内可导，这时对于内每一个确定的值，都对应着f (x)的一个确定的导数，如此一来每一个导数就构成了一个新的函数，这个函数称作 原函数 f (x)的导函数，记作：y'或者f′ (x)。 函数f (x)在它的每一个可导点x。 处都对应着一个唯一确定的数值——导数值f′ (x)，这个对应关系给出了一个定义在f (x)全体可导点的集合上的新函数，称为函数f (x)的导函数，记为f′ (x)。 扩展资 …

2014年12月16日 · The goal is to compute the Fourier series of $g(x)=\log\sin x$ over $[0,\pi]$, or the Fourier series of $h(x)=\log\sin\frac{x}{2}$ over $[0,2\pi]$, or the Fourier series of $f(x)=\log\cos\frac{x}{2}$ over $[-\pi,\pi]$.

2024年12月16日 · Example 34 Evaluate ∫_0^(𝜋/2 ) log⁡sin⁡𝑥 𝑑𝑥 Let I1=∫_0^(𝜋/2 ) 𝑙𝑜𝑔(𝑠𝑖𝑛𝑥) 𝑑𝑥 ∴ I1=∫_0^(𝜋/2) 𝑠𝑖𝑛(𝜋/2−𝑥)𝑑𝑥 I1= ∫_0^(𝜋/2) 𝑙𝑜𝑔(cos⁡𝑥 )𝑑𝑥 Adding (1) and (2) i.e. (1) + (2) I1+ I1=∫_0^(𝜋/2) 〖𝑙𝑜𝑔(sin⁡𝑥 )𝑑𝑥+∫_0^(𝜋/2 ...

2024年12月16日 · log sin⁡𝑥 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. (𝑑(log⁡𝑣))/𝑑𝑥 = 𝑑(cos⁡𝑥. log sin⁡𝑥 )/𝑑𝑥 (𝑑(log⁡𝑣))/𝑑𝑥 . 𝑑𝑣/𝑑𝑣 = 𝑑(cos⁡𝑥.

2024年12月16日 · Example 29 Differentiate 𝑥^sin⁡𝑥 , 𝑥 > 0 𝑤.𝑟.𝑡. 𝑥.Let y = 𝑥^sin⁡𝑥 Taking log both sides log⁡𝑦 = log 𝑥^sin⁡𝑥 𝒍𝒐𝒈⁡𝒚 = 𝒔𝒊𝒏⁡𝒙 . 𝒍𝒐𝒈 𝒙 Differentiating 𝑤.𝑟.𝑡.𝑥 (𝑑(log⁡〖𝑦)〗)/𝑑𝑥 = 𝑑/𝑑𝑥 (sin⁡〖𝑥 log⁡𝑥 〗 ) By product ...

• log (a,x) — \ (\log_a\left (x\right)\) • ln^2 (x), ln (x)^2 — \ (\ln^2\left (x\right)\) 一步一步的极限计算器。 方便的输入和解释！