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恐懼感8964 copyright protection550ＰＥＮＡＮＡOcr3BEIBwu 維尼

554Please respect copyright.ＰＥＮＡＮＡCSgpIJRcSb
8964 copyright protection550ＰＥＮＡＮＡc2bZtBEZFZ 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection550ＰＥＮＡＮＡZjliB1MGWT 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection550ＰＥＮＡＮＡ33WszvY5do 維尼

=2∫eusin(u+a)du… or choose an alternative:554Please respect copyright.ＰＥＮＡＮＡ5cmGYNSFMR
Substitute e√x8964 copyright protection550ＰＥＮＡＮＡHsDc5AifXE 維尼

Now solving:8964 copyright protection550ＰＥＮＡＮＡ44O66KkxmZ 維尼

∫eusin(u+a)du8964 copyright protection550ＰＥＮＡＮＡgpvapfnU76 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection550ＰＥＮＡＮＡW6Md2v1RnY 維尼

First time:8964 copyright protection550ＰＥＮＡＮＡdFxHvWiBeZ 維尼

f=sin(u+a),g′=eu8964 copyright protection550ＰＥＮＡＮＡLBdEChhRQV 維尼

↓ steps↓ steps8964 copyright protection550ＰＥＮＡＮＡpx3mmfC7Qh 維尼

f′=cos(u+a),g=eu:8964 copyright protection550ＰＥＮＡＮＡcnjmFYUhNe 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection550ＰＥＮＡＮＡfm7argz08V 維尼

Second time:8964 copyright protection550ＰＥＮＡＮＡvxirSYI4SE 維尼

f=cos(u+a),g′=eu8964 copyright protection550ＰＥＮＡＮＡWS8YmPI2Tt 維尼

↓ steps↓ steps8964 copyright protection550ＰＥＮＡＮＡshNJwHzrC7 維尼

f′=−sin(u+a),g=eu:8964 copyright protection550ＰＥＮＡＮＡaDK8tEJFpy 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection550ＰＥＮＡＮＡWt3v2u9Evv 維尼

Apply linearity:8964 copyright protection550ＰＥＮＡＮＡeO3EAwkJmF 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection550ＰＥＮＡＮＡ9lR9U4zryG 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection550ＰＥＮＡＮＡYeSqXyC5Op 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection550ＰＥＮＡＮＡpb7fZ5k3cI 維尼

Plug in solved integrals:8964 copyright protection550ＰＥＮＡＮＡn9NMyNCbMy 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection550ＰＥＮＡＮＡ6XWpyn235B 維尼

Undo substitution u=√x:8964 copyright protection550ＰＥＮＡＮＡONeP5FOFxg 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection550ＰＥＮＡＮＡ0781cqzMBn 維尼

The problem is solved:8964 copyright protection550ＰＥＮＡＮＡktZ7wuX1AG 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection550ＰＥＮＡＮＡzeYoyS8eCz 維尼

Rewrite/simplify:8964 copyright protection550ＰＥＮＡＮＡ1HiULnuw7X 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection550ＰＥＮＡＮＡG6amwfkbwU 維尼

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