0-pi-0-2sin-1-rsin-r-dr-d- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 98744 by M±th+et+s last updated on 15/Jun/20 ∫0π∫02sinθ(1+rsinθ)rdrdθ Commented by bemath last updated on 16/Jun/20 =∫π0[12r2+13r3sinθ]02sinθdθ=∫π0(2sin2θ+83sin4θ)dθ=∫π0(1−cos2θ+83(12−12cos2θ)2dθ=∫π0(1−cos2θ+83(14−12cos2θ+14(12+12cos4θ))dθ=∫π0(1−cos2θ+1−43cos2θ+13cos4θ)dθ=[2θ−76sin2θ+112sin4θ]0π=2π Commented by M±th+et+s last updated on 16/Jun/20 thankyousir Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-33207Next Next post: find-lim-x-x-e-x-2-x-1-0-e-t-2-dt- Leave a Reply Cancel replyYour email address will not be published.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.