No Plagiarism!yrQVxP16anUARSX0C2bOposted on PENANA 恐懼感8964 copyright protection492PENANAKppfMJKOWz 維尼
496Please respect copyright.PENANAtdjl8bmCWV
8964 copyright protection492PENANAALXkIin9Og 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection492PENANAIo9sP4xiG0 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection492PENANARbYhp8DRdm 維尼
=2∫eusin(u+a)du… or choose an alternative:496Please respect copyright.PENANAbw52Abwcjo
Substitute e√x8964 copyright protection492PENANAsVG16SQvDd 維尼
Now solving:8964 copyright protection492PENANA0w6p3DCboO 維尼
∫eusin(u+a)du8964 copyright protection492PENANAzQXLoPVBTH 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection492PENANAoq1CwPm3cH 維尼
First time:8964 copyright protection492PENANAmCFazS4YmZ 維尼
f=sin(u+a),g′=eu8964 copyright protection492PENANAKCHFQwPpuG 維尼
↓ steps↓ steps8964 copyright protection492PENANAf2PTrruxPw 維尼
f′=cos(u+a),g=eu:8964 copyright protection492PENANA58Y7puzr8H 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection492PENANAbtk62oUOvN 維尼
Second time:8964 copyright protection492PENANAajTcP7xc3V 維尼
f=cos(u+a),g′=eu8964 copyright protection492PENANAgwldekT7GQ 維尼
↓ steps↓ steps8964 copyright protection492PENANAAFKWWRPFbi 維尼
f′=−sin(u+a),g=eu:8964 copyright protection492PENANAUGaVpmx75A 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection492PENANA7ZBnxjFG4A 維尼
Apply linearity:8964 copyright protection492PENANA7v0c6cpD81 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection492PENANACV2b0dnw1h 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection492PENANAPqUWrAKQRM 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection492PENANAyxXups9dj7 維尼
Plug in solved integrals:8964 copyright protection492PENANAvf3gdpyM3y 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection492PENANAVHlb0O6V3Y 維尼
Undo substitution u=√x:8964 copyright protection492PENANAlU4lIW4wIb 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection492PENANA2eQYmwslXI 維尼
The problem is solved:8964 copyright protection492PENANAVTNWeIb0M8 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection492PENANAzDTsEFEynm 維尼
Rewrite/simplify:8964 copyright protection492PENANA8gsOhKUmAI 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection492PENANAl7FEdEhExe 維尼
18.116.26.90
ns18.116.26.90da2