find-0-1-dt-t-1-t-2-dt-

Question Number 42394 by abdo.msup.com last updated on 24/Aug/18

f i n d \int_{0}^{1} \frac{d t}{t + \sqrt{1 - t^{2}}} d t

Commented by maxmathsup by imad last updated on 25/Aug/18

c h a n g e m e n t t = s i n x g i v e

I = \int_{0}^{\frac{π}{2}} \frac{c o s x d x}{s i n x + c o s x} d x = \int_{0}^{\frac{π}{2}} \frac{d x}{t a n x + 1}

I =_{t a n x = u} \int_{0}^{+ \infty} \frac{1}{1 + u} \frac{d u}{1 + u^{2}} = \int_{0}^{\infty} \frac{d u}{(u + 1) (u^{2} + 1)} l e t d e c o m p o s e

F (u) = \frac{1}{(u + 1) (u^{2} + 1)} \Rightarrow F (u) = \frac{a}{u + 1} + \frac{b u + c}{u^{2} + 1}

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

{l i m}_{u \to + \infty} u F (u) = 0 = a + b \Rightarrow b = - \frac{1}{2} \Rightarrow F (u) = \frac{1}{2 (u + 1)} + \frac{- \frac{1}{2} u + c}{u^{2} + 1}

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

I = \int_{0}^{\infty} F (u) d u = \frac{1}{2} \int_{0}^{\infty} \frac{d u}{u + 1} - \frac{1}{4} \int_{0}^{\infty} \frac{2 u - 2}{u^{2} + 1} d u

= {[\frac{1}{2} l n ∣ u + 1 ∣ - \frac{1}{4} l n (u^{2} + 1)]}_{0}^{+ \infty} + \frac{1}{2} {[a r c t a n u]}_{0}^{+ \infty}

= \frac{1}{2} {[l n ∣ \frac{u + 1}{\sqrt{u^{2} + 1}} ∣]}_{0}^{+ \infty} + \frac{π}{4} = 0 + \frac{π}{4} = \frac{π}{4} .

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Aug/18

t = s i n α d t = c o s α d α

\int_{0}^{\frac{Π}{2}} \frac{c o s α}{c o s α + s i n α} d α

= \frac{1}{2} \int_{0}^{\frac{Π}{2}} \frac{c o s α + s i n α + c o s α - s i n α}{c o s α + s i n α} d α

= \frac{1}{2} \int_{0}^{\frac{Π}{2}} d α + \frac{1}{2} \int_{0}^{\frac{Π}{2}} \frac{d (s i n α + c o s α)}{c o s α + s i n α}

= \frac{1}{2} (\frac{Π}{2}) + \frac{1}{2} ∣ l n (c o s α + s i n α) ∣_{0}^{\frac{Π}{2}}

= \frac{Π}{4} + \frac{1}{2} {l n (1) - l n 1}

= \frac{Π}{4}