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No Plagiarism!SdLgMgLAi4C6ccYV0ohKposted on PENANA 恐懼感8964 copyright protection513PENANA3IqeYGsSAI 維尼
517Please respect copyright.PENANASdRW7eXlLa
8964 copyright protection513PENANACggS9JUfNY 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection513PENANAW154eHBsba 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection513PENANAttJvlNnQ37 維尼
=2∫eusin(u+a)du… or choose an alternative:517Please respect copyright.PENANAbxOCGHnumC
Substitute e√x8964 copyright protection513PENANA9wnapjZySF 維尼
Now solving:8964 copyright protection513PENANAb8nEs7ZmBM 維尼
∫eusin(u+a)du8964 copyright protection513PENANAy4zEuXGl55 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection513PENANAbXHBqNYp0B 維尼
First time:8964 copyright protection513PENANAINVLKF0KcA 維尼
f=sin(u+a),g′=eu8964 copyright protection513PENANAzArqhbmz6M 維尼
↓ steps↓ steps8964 copyright protection513PENANAXOaedLFYZi 維尼
f′=cos(u+a),g=eu:8964 copyright protection513PENANAdv15wS45KJ 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection513PENANAQj486SPDxy 維尼
Second time:8964 copyright protection513PENANAOUkLLYvIfr 維尼
f=cos(u+a),g′=eu8964 copyright protection513PENANApXOk02TGeF 維尼
↓ steps↓ steps8964 copyright protection513PENANAHONlGSKo8j 維尼
f′=−sin(u+a),g=eu:8964 copyright protection513PENANAplzxT08VXp 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection513PENANAaR0HQApafA 維尼
Apply linearity:8964 copyright protection513PENANAJvzbSb93Rs 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection513PENANA8otTfk1boq 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection513PENANAJdG7adEvKz 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection513PENANAae4Gtbm2SH 維尼
Plug in solved integrals:8964 copyright protection513PENANATuzzQbzJVb 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection513PENANALnFmiQyLsy 維尼
Undo substitution u=√x:8964 copyright protection513PENANAtCiydwviSR 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection513PENANAcYr0hYreRe 維尼
The problem is solved:8964 copyright protection513PENANAqxj3xzu0Nv 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection513PENANAMWvLi1DkgY 維尼
Rewrite/simplify:8964 copyright protection513PENANAeZtvq1SXqJ 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection513PENANAaNPTD4qP44 維尼
216.73.216.39
ns216.73.216.39da2