find-I-0-pi-2-1-sin-cos-d-

Question Number 31073 by abdo imad last updated on 02/Mar/18

f i n d I = \int_{0}^{\frac{π}{2}} \frac{1 - s i n θ}{c o s θ} d θ .

Commented by prof Abdo imad last updated on 03/Mar/18

t h e c h . t a n (\frac{θ}{2}) = x g i v e

I = \int_{0}^{1} \frac{1 - \frac{2 x}{1 + x^{2}}}{\frac{1 - x^{2}}{1 + x^{2}}} \frac{2 d x}{1 + x^{2}} = 2 \int_{0}^{1} \frac{{(x - 1)}^{2}}{(1 - x^{2}) (1 + x^{2})} d x

= 2 \int_{0}^{1} \frac{1 - x}{(1 + x) (1 + x^{2})} d x l e t d e c o m p o s e t h e f r s c t i o n

F (x) = \frac{1 - x}{(1 + x) (1 + x^{2})} = \frac{a}{1 + x} + \frac{b x + c}{1 + x^{2}}

a = {l i m}_{x \to - 1} (x + 1) F (x) = \frac{2}{2} = 1

{l i m}_{x \to + \infty} x F (x) = 0 = a + b \Rightarrow b = - a = - 1 \Rightarrow

F (x) = \frac{1}{1 + x} + \frac{- x + c}{1 + x^{2}}

F (0) = 1 = 1 + c \Rightarrow c = 0 s o F (x) = \frac{1}{1 + x} - \frac{x}{1 + x^{2}} \Rightarrow

I = 2 \int_{0}^{1} \frac{d x}{1 + x} d x + \int_{0}^{1} \frac{2 x}{1 + x^{2}} d x p

= 2 {[l n ∣ 1 + x ∣]}_{0}^{1} - {[l n (1 + x^{2})]}_{0}^{1} = 2 l n (2) - l n (2) \Rightarrow

I = l n (2) .

Answered by Joel578 last updated on 02/Mar/18

I = \lim_{t \to \frac{π}{2}} (\int_{0}^{t} \sec x - \tan x d x)

= \lim_{t \to \frac{π}{2}} {[\ln (\sec x + \tan x) \cos x]}_{0}^{t}

= \lim_{t \to \frac{π}{2}} \ln ((\sec t + \tan t) \cos t) - \ln (1 + 0)

= \lim_{t \to \frac{π}{2}} \ln ((\sec t + \tan t) \cos t)

= \lim_{t \to \frac{π}{2}} \ln (1 + \tan t \cos t) = \ln (1 + \lim_{t \to \frac{π}{2}} \tan t \cos t)

= \ln (1 + \lim_{t \to \frac{π}{2}} \frac{\cos t}{\cot t})

= \ln (1 + \lim_{t \to \frac{π}{2}} \frac{- \sin t}{- {cosec}^{2} t}) = \ln 2

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