J-d-tan-cot-sec-csc- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 129173 by bramlexs22 last updated on 13/Jan/21 J=∫dθtanθ+cotθ+secθ+cscθ? Answered by liberty last updated on 13/Jan/21 J=∫dθsinθ+1cosθ+cosθ+1sinθ=∫cosθsinθdθsin2θ+sinθ+cos2θ+cosθJ=∫sinθcosθdθ1+sinθ+cosθ=∫2sin(θ2)cos(θ2)cosθ2sin(θ2)cos(θ2)+2cos2(θ2)dθJ=∫sin(θ2)cosθsin(θ2)+cos(θ2)dθJ=∫sin(θ2)(cos(θ2)−sin(θ2))dθJ=∫12sinθ−(12−12cosθ)dθJ=−12cosθ−12θ+12sinθ+C― Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: 1-y-2-dx-1-x-2-dy-0-y-0-3-2-Next Next post: Question-63639 Leave a Reply Cancel replyYour email address will not be published.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.