Please suggest improvements or revisions to this solution. Calculate integral_0^
ID: 3027898 • Letter: P
Question
Please suggest improvements or revisions to this solution. Calculate integral_0^pi/2 {tan x}/tan x dx, where {u} denotes u - [u]. We show that integral_0^pi/2 {tan x}/tan x dx = pi/2 - 1/2 ln (sin h pi/pi). Because f(x) = tan(x) is a continuous, increasing function from [0, pi/2) onto [0, infinity), for each 11011 negative integer k, there exists a unique b_k in [0, pi/22) such that tan(b_k) = k. integral_b0^b1 {tan x}/tan x dx = integral_0^b1 (tan x) - 0/tan x dx = integral_0^b1 1 dx -= b_1 and when k greaterthaorequalto 1, we have integral_bk^b_k + 1 {tan x}/tan x dx = integral_bk^b_k+ 1 (tan x) - k/tan x dx = integral_b_k^b_k+1 1 - k cos x/sin dx = b_k + 1 - b_k - k [ln sin (b_k - 1) - ln sin sin(b_k)] Adding equation (2) using integer values of k from 0 to n and using sin(b_k) = k/Squareroot k^2 + 1 yelds, for all integers n greaterthanorequalto 1, integral_0^b-n + 1 {tan x}/tan dx dx = b_n + 1 - n ln sin (b_n + 1) + sigma_k = 1^n ln sin (b_k) = b_n + 1 - n ln (n + 1/Squareroot (n + 1)^2 + 1) + ln (Product_k = 1 l/Squareroot k^2 + 1) = b_n + 1 - n [1/2(n + 1)^2 + O (1/(n + 1)^4)] - 1/2 ln (Product_k = 1^n (1 + 1/k^2)) where the first of these equations may also be proved by induction and we used (1 + z)^-1/2 = 1 - 1/2 z + O(z^2) with z = 1/(n + 1)^2 + O(z^2) with z = 1/(n + 1)^2 and ln (1 - t) = t - O(t^2) with t = 1/2 n + O(z^2) when rewriting the second term in the second of these equations. Letting n rightarrow infintiy and using the result that lim_n rightarrow ifnitniy which can be shown by applying the Squeeze Principle to 0 where we used the Euler product expansion.Explanation / Answer
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