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No Plagiarism!k0kRpupp0Ca3pLfCWWyUposted on PＥNAＮＡ 恐懼感8964 copyright protection469ＰＥＮＡＮＡ41i0qxLPYH 維尼
473Please respect copyright.ＰＥＮＡＮＡylwY6WiBjB
8964 copyright protection469ＰＥＮＡＮＡmAO7cevaGP 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection469ＰＥＮＡＮＡszzPRRoDoU 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection469ＰＥＮＡＮＡIzRvZc4h7P 維尼
=2∫eusin(u+a)du… or choose an alternative:473Please respect copyright.ＰＥＮＡＮＡt865YwpHw4
Substitute e√x8964 copyright protection469ＰＥＮＡＮＡHkkTRfiqiz 維尼
Now solving:8964 copyright protection469ＰＥＮＡＮＡpCde7BCq7n 維尼
∫eusin(u+a)du8964 copyright protection469ＰＥＮＡＮＡd0rjrZ5aQG 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection469ＰＥＮＡＮＡ7d10VT4Nyi 維尼
First time:8964 copyright protection469ＰＥＮＡＮＡL4avWfr5zY 維尼
f=sin(u+a),g′=eu8964 copyright protection469ＰＥＮＡＮＡ3AAdnwBTeL 維尼
↓ steps↓ steps8964 copyright protection469ＰＥＮＡＮＡZvUNFmlphZ 維尼
f′=cos(u+a),g=eu:8964 copyright protection469ＰＥＮＡＮＡsolfWPZnBT 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection469ＰＥＮＡＮＡ1SyiAFBmrT 維尼
Second time:8964 copyright protection469ＰＥＮＡＮＡ06kTRgSMhK 維尼
f=cos(u+a),g′=eu8964 copyright protection469ＰＥＮＡＮＡOa7bwVZBqS 維尼
↓ steps↓ steps8964 copyright protection469ＰＥＮＡＮＡd8MNuk9enq 維尼
f′=−sin(u+a),g=eu:8964 copyright protection469ＰＥＮＡＮＡupUETYjMd2 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection469ＰＥＮＡＮＡ1NHVlt5Pj7 維尼
Apply linearity:8964 copyright protection469ＰＥＮＡＮＡ0Ovu4T1BjM 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection469ＰＥＮＡＮＡ6lrmzaOmUu 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection469ＰＥＮＡＮＡ5WDbTAsNqw 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection469ＰＥＮＡＮＡfg8rqYeq7t 維尼
Plug in solved integrals:8964 copyright protection469ＰＥＮＡＮＡ6zWu5mn72f 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection469ＰＥＮＡＮＡnji0XGlFr8 維尼
Undo substitution u=√x:8964 copyright protection469ＰＥＮＡＮＡ94hkbLLmyq 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection469ＰＥＮＡＮＡ0hLuq4AGQQ 維尼
The problem is solved:8964 copyright protection469ＰＥＮＡＮＡsCBnolGoXh 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection469ＰＥＮＡＮＡLbWPlMY91L 維尼
Rewrite/simplify:8964 copyright protection469ＰＥＮＡＮＡ3fD9awjo52 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection469ＰＥＮＡＮＡSkkltr4yss 維尼
3.141.7.31
ns3.141.7.31da2
