calculate-0-pi-2-ln-tan-x-dx-0-sin-2-x-x-2-dx-ln-sin-x-dx- Tinku Tara November 16, 2023 Integration 0 Comments FacebookTweetPin Question Number 200254 by mnjuly1970 last updated on 16/Nov/23 calculate…Ω=∫∫0π2ln(tan(x))dx∫0∞sin2(x)x2dxln(sin(x))dx=? Answered by Mathspace last updated on 16/Nov/23 itsatrikyintegralwehave∫0π2ln(tanx)dx=∫0π2ln(sinxcosx)dx=∫0π2ln(sinx)dx−∫0π2ln(cosx)dx=0(equal)∫0∞sin2xx2dx=[−1xsin2x]0∞−∫0∞(−1x)2sinxcosxdx=0+∫0∞sin(2x)xdx(2x=t)=∫0∞sintt2dt2=∫0∞sinttdt=π2⇒I=∫0π2ln(sinx)dx=−π2ln2 Commented by Calculusboy last updated on 16/Nov/23 thankssir Commented by mnjuly1970 last updated on 17/Nov/23 thanksalotsir Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-200250Next Next post: Question-200253 Leave a Reply Cancel replyYour email address will not be published.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.