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No Plagiarism!zEEQcCiAeCXNPVAM5lj2posted on PＥNAＮＡ 恐懼感8964 copyright protection422ＰＥＮＡＮＡZQDwcjWwyT 維尼
426Please respect copyright.ＰＥＮＡＮＡJmvPazGKAe
8964 copyright protection422ＰＥＮＡＮＡ7ZByCvRYQ3 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection422ＰＥＮＡＮＡaUdKq7LHWS 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection422ＰＥＮＡＮＡWqw2yIkcKQ 維尼
=2∫eusin(u+a)du… or choose an alternative:426Please respect copyright.ＰＥＮＡＮＡGk0te0oXW7
Substitute e√x8964 copyright protection422ＰＥＮＡＮＡ4vEEsgl3IR 維尼
Now solving:8964 copyright protection422ＰＥＮＡＮＡJV3GAZ0d2h 維尼
∫eusin(u+a)du8964 copyright protection422ＰＥＮＡＮＡeCHKg2ZXHw 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection422ＰＥＮＡＮＡy7bTDz6bW8 維尼
First time:8964 copyright protection422ＰＥＮＡＮＡr7bzlYkVPP 維尼
f=sin(u+a),g′=eu8964 copyright protection422ＰＥＮＡＮＡA6Aiiv0OHH 維尼
↓ steps↓ steps8964 copyright protection422ＰＥＮＡＮＡvHnYr2GBXU 維尼
f′=cos(u+a),g=eu:8964 copyright protection422ＰＥＮＡＮＡySjdvXnlR5 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection422ＰＥＮＡＮＡpbh1sSL0zE 維尼
Second time:8964 copyright protection422ＰＥＮＡＮＡH58MpHWOhT 維尼
f=cos(u+a),g′=eu8964 copyright protection422ＰＥＮＡＮＡTIi84Q7yMT 維尼
↓ steps↓ steps8964 copyright protection422ＰＥＮＡＮＡAsFIk86b3n 維尼
f′=−sin(u+a),g=eu:8964 copyright protection422ＰＥＮＡＮＡmGlToTFnWc 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection422ＰＥＮＡＮＡ7IhTKCpTXZ 維尼
Apply linearity:8964 copyright protection422ＰＥＮＡＮＡMGFYOgKdDW 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection422ＰＥＮＡＮＡTp8jGzCZlQ 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection422ＰＥＮＡＮＡAykjoQtHgJ 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection422ＰＥＮＡＮＡFmT4LhM5Nw 維尼
Plug in solved integrals:8964 copyright protection422ＰＥＮＡＮＡrSuva0UF7L 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection422ＰＥＮＡＮＡgOCOCWnlXF 維尼
Undo substitution u=√x:8964 copyright protection422ＰＥＮＡＮＡHZKi8YOG15 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection422ＰＥＮＡＮＡJqkWTnjhU0 維尼
The problem is solved:8964 copyright protection422ＰＥＮＡＮＡ29elphMP6Z 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection422ＰＥＮＡＮＡ5KUeqOCEqs 維尼
Rewrite/simplify:8964 copyright protection422ＰＥＮＡＮＡSFSgMoUZAC 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection422ＰＥＮＡＮＡR9RG5NQ2Cb 維尼
3.144.118.174
ns3.144.118.174da2
