No Plagiarism!UWAC7jKWtWoc2jgll9GLposted on PENANA 恐懼感8964 copyright protection564PENANAhHHp8VJ2mH 維尼
568Please respect copyright.PENANAdbnJy8LAxx
8964 copyright protection564PENANAXVOSsSCHeD 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection564PENANAtuqSoNizm7 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection564PENANAIJ7Q4M8yP5 維尼
=2∫eusin(u+a)du… or choose an alternative:568Please respect copyright.PENANAWxzyOOJYvm
Substitute e√x8964 copyright protection564PENANA4Nrg2TvCMH 維尼
Now solving:8964 copyright protection564PENANA6TxNRhEQ2J 維尼
∫eusin(u+a)du8964 copyright protection564PENANAtEA3fiovxp 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection564PENANA4i82CRyYsy 維尼
First time:8964 copyright protection564PENANAyBVH2ImVRv 維尼
f=sin(u+a),g′=eu8964 copyright protection564PENANAaCt9sfYXIM 維尼
↓ steps↓ steps8964 copyright protection564PENANABhnsqiBkHI 維尼
f′=cos(u+a),g=eu:8964 copyright protection564PENANASfxLuq6I8p 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection564PENANAO3bdAQO5zO 維尼
Second time:8964 copyright protection564PENANARaTgpFuZ8N 維尼
f=cos(u+a),g′=eu8964 copyright protection564PENANAevrJYisPRL 維尼
↓ steps↓ steps8964 copyright protection564PENANA9pvDcqpV2I 維尼
f′=−sin(u+a),g=eu:8964 copyright protection564PENANAekdtEzHfZ2 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection564PENANAWEK645jiJo 維尼
Apply linearity:8964 copyright protection564PENANABV6CVuX6Tk 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection564PENANAbC91zNRIed 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection564PENANA6L6FSNBQJP 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection564PENANABXdWV5p8CA 維尼
Plug in solved integrals:8964 copyright protection564PENANACqkLOua5v3 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection564PENANAuCB9707Hfr 維尼
Undo substitution u=√x:8964 copyright protection564PENANAeXnDnt1JIS 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection564PENANAmc8zicc0qy 維尼
The problem is solved:8964 copyright protection564PENANAYiILiWLmXe 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection564PENANA33Bl7rhIlE 維尼
Rewrite/simplify:8964 copyright protection564PENANAs7qELxfDzF 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection564PENANA8UnGEHSCed 維尼
216.73.216.228
ns216.73.216.228da2