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No Plagiarism!qicySthAgr4Mp0ujvAvrposted on PＥNAＮＡ 恐懼感8964 copyright protection458ＰＥＮＡＮＡ8Qtw0nXwZC 維尼
462Please respect copyright.ＰＥＮＡＮＡHWJFwRZYhF
8964 copyright protection458ＰＥＮＡＮＡ3kVFRhNdCj 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection458ＰＥＮＡＮＡjl1e3t2Kry 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection458ＰＥＮＡＮＡRKmCoNcnHn 維尼
=2∫eusin(u+a)du… or choose an alternative:462Please respect copyright.ＰＥＮＡＮＡU5vbUHc4ZM
Substitute e√x8964 copyright protection458ＰＥＮＡＮＡMXgZ7etbwi 維尼
Now solving:8964 copyright protection458ＰＥＮＡＮＡTmHwgsyHuZ 維尼
∫eusin(u+a)du8964 copyright protection458ＰＥＮＡＮＡO7nPpT3v7S 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection458ＰＥＮＡＮＡKNAVoetXYF 維尼
First time:8964 copyright protection458ＰＥＮＡＮＡScPIsc46pO 維尼
f=sin(u+a),g′=eu8964 copyright protection458ＰＥＮＡＮＡrqPRV5ElV1 維尼
↓ steps↓ steps8964 copyright protection458ＰＥＮＡＮＡNS4p26INjr 維尼
f′=cos(u+a),g=eu:8964 copyright protection458ＰＥＮＡＮＡpS9p9n09fh 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection458ＰＥＮＡＮＡLjiwctKtQM 維尼
Second time:8964 copyright protection458ＰＥＮＡＮＡVGZqubmRr1 維尼
f=cos(u+a),g′=eu8964 copyright protection458ＰＥＮＡＮＡBy222XqvnJ 維尼
↓ steps↓ steps8964 copyright protection458ＰＥＮＡＮＡLJXm75ASQT 維尼
f′=−sin(u+a),g=eu:8964 copyright protection458ＰＥＮＡＮＡsDnQwTcpit 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection458ＰＥＮＡＮＡVQt1Apg302 維尼
Apply linearity:8964 copyright protection458ＰＥＮＡＮＡ2qLvxAsowH 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection458ＰＥＮＡＮＡXSeXXfFNvV 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection458ＰＥＮＡＮＡGMEhpzE3VO 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection458ＰＥＮＡＮＡgrXHWCL45t 維尼
Plug in solved integrals:8964 copyright protection458ＰＥＮＡＮＡKXAypI3ufA 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection458ＰＥＮＡＮＡDFiXXR42Ds 維尼
Undo substitution u=√x:8964 copyright protection458ＰＥＮＡＮＡ6jNEECugHI 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection458ＰＥＮＡＮＡrdZHs30Umy 維尼
The problem is solved:8964 copyright protection458ＰＥＮＡＮＡv5LbrIrLrk 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection458ＰＥＮＡＮＡHAUL4cMfmj 維尼
Rewrite/simplify:8964 copyright protection458ＰＥＮＡＮＡORJ4T9KPyS 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection458ＰＥＮＡＮＡdjHyLtNHY9 維尼
3.15.0.42
ns3.15.0.42da2
