
log (sinx) - Wolfram|Alpha
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Example 34 - Integration of log (sin x) from 0 to pi/2 - Teachoo
2024年12月16日 · Example 34 Evaluate ∫_0^ (𝜋/2 ) logsin𝑥 𝑑𝑥 Let I1=∫_0^ (𝜋/2 ) 𝑙𝑜𝑔 (𝑠𝑖𝑛𝑥) 𝑑𝑥 ∴ I1=∫_0^ (𝜋/2) 𝑠𝑖𝑛 (𝜋/2−𝑥)𝑑𝑥 I1= ∫_0^ (𝜋/2) 𝑙𝑜𝑔 (cos𝑥 )𝑑𝑥 Adding (1) and (2) i.e. (1) + (2) I1+ I1=∫_0^ (𝜋/2) 〖𝑙𝑜𝑔 (sin𝑥 )𝑑𝑥+∫_0 ...
calculus - Computing the integral of $\log (\sin x)
How to compute the following integral? $$\int\log (\sin x)\,dx$$ Motivation: Since $\log (\sin x)'=\cot x$, the antiderivative $\int\log (\sin x)\,dx$ has the nice property $F'' (x)=\cot x$.
derivative of log(sinx) - Symbolab
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不定积分计算器 : log (sin (x)) - 数字帝国
不定积分计算器 可以用分析整合的方法,计算出一个给定变量的函数的不定积分(原函数)。 它也可以画出函数和它的原函数的图像。 请注意,计算的不定积分属于一类函数F (x) C,其中C是任意常数。 不定积分计算器解析表达式,应用积分法则并化简最终结果。 因此,积分计算的最终结果可能与常数的预期结果不同。 不定积分计算器可以用分析整合的方法,计算出一个给定变量的函数的不定积分(原函数)。 它也可以画出函数和它的原函数的图像。 请注意,计算的不定积 …
计算器是如何计算sin、log等函数的?底层运算逻辑是什么? - 知乎
sin log要么是把结果预先存储在计算器rom里,要么是用 泰勒展开 转换成多项式计算. 确定定义域之后,可以用泰勒级数,不过泰勒级数在哪一点展开呢? 方法二:预先计算出一些点,查表找最近的点展开;点足够密集时,一阶近似就能满足精度要求。 在一些低功耗的定制芯片里也会用CORDIC算法,和泰勒级数相比优点就是不需要乘除法(除了乘除2的幂以外)。 在 我以前的文章 里提到过,通用芯片编程的时候如果已经有了乘法器,那就放心大胆用,但是设计定制芯片 …
求解 log (sin (x)) | Microsoft Math Solver
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. ...
Log Sine Function - from Wolfram MathWorld
4 天之前 · The log sine function, also called the logsine function, is defined by S_n=int_0^pi [ln (sinx)]^ndx. (1) The first few cases are given by S_1 = -piln2 (2) S_2 = 1/ (12)pi^3+pi (ln2)^2 (3) S_3 = -1/4pi^3ln2-pi (ln2)^3-3/2pizeta (3), (4) where zeta (z) is the Riemann zeta function.
solve log (x) = sin (x) - Wolfram|Alpha
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Compute the fourier coefficients, and series for $\\log(\\sin(x))$
2014年12月16日 · log(sin(x)) = − log(2) − ∞ ∑ k = 1cos(2kx) k. Notice that an is not so difficult to compute, if one performs a round of integration by parts, bringing the logarithm into a cotangent and the cosine into a sine. The only slightly tricky part is to compute a0, but there is a nice proof through Riemann sums.