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恐懼感8964 copyright protection382ＰＥＮＡＮＡ1HghcvxrMU 維尼

386Please respect copyright.ＰＥＮＡＮＡUbiZsdns9g
8964 copyright protection382ＰＥＮＡＮＡefRRGWMkIi 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection382ＰＥＮＡＮＡNXu4jgzYPP 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection382ＰＥＮＡＮＡLEfsVVoUAE 維尼

=2∫eusin(u+a)du… or choose an alternative:386Please respect copyright.ＰＥＮＡＮＡUqtlZoAgUK
Substitute e√x8964 copyright protection382ＰＥＮＡＮＡgnB9Z2UmAr 維尼

Now solving:8964 copyright protection382ＰＥＮＡＮＡWvsoCm4Dty 維尼

∫eusin(u+a)du8964 copyright protection382ＰＥＮＡＮＡ8PCjhdc9Yk 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection382ＰＥＮＡＮＡV7cWtItsUE 維尼

First time:8964 copyright protection382ＰＥＮＡＮＡlFERuiYDrl 維尼

f=sin(u+a),g′=eu8964 copyright protection382ＰＥＮＡＮＡf4YAktjMyy 維尼

↓ steps↓ steps8964 copyright protection382ＰＥＮＡＮＡLHB7ENIv2D 維尼

f′=cos(u+a),g=eu:8964 copyright protection382ＰＥＮＡＮＡXIvTWCASsm 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection382ＰＥＮＡＮＡztLHByqO4U 維尼

Second time:8964 copyright protection382ＰＥＮＡＮＡomifjRgjkl 維尼

f=cos(u+a),g′=eu8964 copyright protection382ＰＥＮＡＮＡuhCZsOCzQE 維尼

↓ steps↓ steps8964 copyright protection382ＰＥＮＡＮＡCSpfugzay3 維尼

f′=−sin(u+a),g=eu:8964 copyright protection382ＰＥＮＡＮＡa2eclXotan 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection382ＰＥＮＡＮＡfRLMT0NDFt 維尼

Apply linearity:8964 copyright protection382ＰＥＮＡＮＡEBIdi6gYLr 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection382ＰＥＮＡＮＡEPifsP2JIU 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection382ＰＥＮＡＮＡVTHxSgMq45 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection382ＰＥＮＡＮＡmNYjMotalD 維尼

Plug in solved integrals:8964 copyright protection382ＰＥＮＡＮＡJ2frXa5GXU 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection382ＰＥＮＡＮＡ9kLj9rVihk 維尼

Undo substitution u=√x:8964 copyright protection382ＰＥＮＡＮＡojDxSSc4AS 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection382ＰＥＮＡＮＡoHTgBZRVYf 維尼

The problem is solved:8964 copyright protection382ＰＥＮＡＮＡ9zV9rMZJ5U 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection382ＰＥＮＡＮＡVsjf5kn96o 維尼

Rewrite/simplify:8964 copyright protection382ＰＥＮＡＮＡ13uRqJOuS4 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection382ＰＥＮＡＮＡHvV1WmjAYB 維尼

15.158.61.6