0-xsin-4x-9-4x-2-dx- Tinku Tara June 4, 2023 Others 0 Comments FacebookTweetPin Question Number 146621 by 777316 last updated on 14/Jul/21 ∫0∞xsin(4x)9+4x2dx= Answered by mathmax by abdo last updated on 14/Jul/21 Ψ=∫0∞xsin(4x)4x2+9⇒Ψ=2x=3z∫0∞3zsin(43z2)2.9(1+z2)32dz=14∫0∞zsin(6z)z2+1dz=18∫−∞+∞zsin(6z)z2+1dz=18Im(∫−∞+∞ze6izz2)1dz)letφ(z)=ze6izz2+1⇒φ(z)=ze6iz(z−i)(z+i)residusth.⇒∫Rφ(z)dz=2iπRes(φ,i)Res(φ,i)=ie−62i⇒∫Rφ(z)dz=2iπ.ie−62i=iπe6⇒Ψ=18×πe6=π8e6 Answered by KINMATICS last updated on 14/Jul/21 ∫xsin4x9+4x2dx=∫sin4x.x(2x)2−(3i)2dx∫xsin4x9+4x2dx=∫sin4x.14(1(2x+3i)+12x−3i)dx∫xsin4x9+4x2dx=12∫sin4x(4x+6i)dx+12∫sin4x4x−6idxFor∫sin4x4x+6idx,let4x+6i=u,4x=u−6i,dx=du4for∫sin4x4x−6idx,let4x−6i=y,4x=y+6i,dx=dy4⇒∫sin4xdx4x2+9=18∫sin(u−6i)udu+18∫sin(y+6i)ydyI=18∫sin(u)cos(6i)−cos(u)sin(6i)udu+18∫sin(y)cos(6i)+cos(y)sin(6i)ydyI=cos(6i)8∫sin(u)udu−sin(6i)8∫cos(u)udu+cos(6i)8∫sin(y)ydy+sin(6i)8∫cos(y)ydyI=cos(6i)8Si(u)−sin(6i)8Ci(u)+cos(6i)8Si(y)+sin(6i)8Ci(y)+CI=cos(6i)8Si(4x+6i)−sin(6i)8Ci(4x+6i)cos(6i)8Si(4x−6i)+sin(6i)8Ci(4x−6i)+CBy:KinTomcontactwww.hppt://kintom077∂gmail.com Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: solve-x-R-x-x-2-x-6-note-x-max-q-Z-q-x-Next Next post: Question-81091 Leave a Reply Cancel replyYour email address will not be published.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.