Question-150413 Tinku Tara June 4, 2023 None 0 Comments FacebookTweetPin Question Number 150413 by saly last updated on 12/Aug/21 Commented by MJS_new last updated on 12/Aug/21 inthiscasewemusttakethelongwayhome…(1)arctanr=−i2(ln(1+ir)−ln(1−ir))⇒∫arctansinxdx==−i2(∫ln(1+isinx)dx−∫ln(1−isinx)dx)(2)sinα=eiα−e−iα2i⇒[similarforthe2ndintegral]∫ln(1+isinx)dx==∫ln(1+eix2−12eix)dx(3)t=ix⇔x=−it→dx=−idt⇒∫ln(1+eix2−12eix)dx==iln2∫dt−i∫ln(2+et−e−t)dt(4)byparts⇒∫ln(2+et−e−t)dt==tln(2+et−e−t)−∫t(e2t+1)e2t−2et−1dt(5)u=et⇔t=lnu→dt=duu⇒∫t(e2t+1)e2t−2et−1dt==∫(u2+1)lnuu(u2−2u−1)du==2∫ulnuu2−2u−1du−2∫lnuu2−2u−1du−∫lnuuduandIhopeyoucandotheseyourself Commented by saly last updated on 12/Aug/21 Thanksir Commented by MJS_new last updated on 12/Aug/21 you′rewelcome. Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: e-2dx-xlnx-Next Next post: Prove-that-z-1-z-2-z-3-z-n-z-1-z-2-z-3-z-n- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![in this case we must take the long way home... (1) arctan r =−(i/2)(ln (1+ir) −ln (1−ir)) ⇒ ∫arctan sin x dx= =−(i/2)(∫ln (1+i sin x) dx−∫ln (1−i sin x) dx) (2) sin α =((e^(iα) −e^(−iα) )/(2i)) ⇒ [similar for the 2^(nd) integral] ∫ln (1+i sin x) dx= =∫ln (1+(e^(ix) /2)−(1/(2e^(ix) ))) dx (3) t=ix ⇔ x=−it → dx=−idt ⇒ ∫ln (1+(e^(ix) /2)−(1/(2e^(ix) ))) dx= =iln 2∫dt −i∫ln (2+e^t −e^(−t) ) dt (4) by parts ⇒ ∫ln (2+e^t −e^(−t) ) dt= =tln (2+e^t −e^(−t) ) −∫((t(e^(2t) +1))/(e^(2t) −2e^t −1))dt (5) u=e^t ⇔ t=ln u → dt=(du/u) ⇒ ∫((t(e^(2t) +1))/(e^(2t) −2e^t −1))dt= =∫(((u^2 +1)ln u)/(u(u^2 −2u−1)))du= =2∫((uln u)/(u^2 −2u−1))du−2∫((ln u)/(u^2 −2u−1))du−∫((ln u)/u)du and I hope you can do these yourself](https://www.tinkutara.com/question/Q150417.png)
