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Question Number 43683 by Raj Singh last updated on 14/Sep/18

Commented by Meritguide1234 last updated on 14/Sep/18

Commented by maxmathsup by imad last updated on 15/Sep/18
$let I = ∫ (√(cotanx))dx ⇒ I = ∫ (dx/(√(tanx))) changement (√(tanx))=t give tanx =t^2 ⇒x =arctan(t^2 ) ⇒ dx =((2t)/(1+t^4 )) dt ⇒ I = ∫ ((2t)/((1+t^4 )t))dt =2 ∫ (dt/(1+t^4 )) but we have proved that ∫ (dt/(1+t^4 )) =(1/(4(√2)))ln∣((t^2 −(√2)t +1)/(t^2 +(√2)t +1))∣ +(1/(2(√2))){ arctan(t(√2)−1) +arctan(t(√2)+1)} ⇒ I = (1/(2(√2)))ln∣ ((tanx −(√(2tanx))+1)/(tanx +(√(2tanx)) +1))∣ +(1/(√2)){ arctan((√(2tanx))−1)+arctan((√(2tanx)) +1)}$
$l e t I = \int \sqrt{c o t a n x} d x \Rightarrow I = \int \frac{d x}{\sqrt{t a n x}} c h a n g e m e n t \sqrt{t a n x} = t g i v e$ $t a n x = t^{2} \Rightarrow x = a r c t a n (t^{2}) \Rightarrow d x = \frac{2 t}{1 + t^{4}} d t \Rightarrow I = \int \frac{2 t}{(1 + t^{4}) t} d t$ $= 2 \int \frac{d t}{1 + t^{4}} b u t w e h a v e p r o v e d t h a t$ $\int \frac{d t}{1 + t^{4}} = \frac{1}{4 \sqrt{2}} l n ∣ \frac{t^{2} - \sqrt{2} t + 1}{t^{2} + \sqrt{2} t + 1} ∣ + \frac{1}{2 \sqrt{2}} {a r c t a n (t \sqrt{2} - 1) + a r c t a n (t \sqrt{2} + 1)} \Rightarrow$ $I = \frac{1}{2 \sqrt{2}} l n ∣ \frac{t a n x - \sqrt{2 t a n x} + 1}{t a n x + \sqrt{2 t a n x} + 1} ∣ + \frac{1}{\sqrt{2}} {a r c t a n (\sqrt{2 t a n x} - 1) + a r c t a n (\sqrt{2 t a n x} + 1)}$
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