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恐懼感8964 copyright protection515ＰＥＮＡＮＡfKEIj2x1pg 維尼

519Please respect copyright.ＰＥＮＡＮＡ3NegHblxDH
8964 copyright protection515ＰＥＮＡＮＡOjZV31OUJY 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection515ＰＥＮＡＮＡodVFkxQnPY 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection515ＰＥＮＡＮＡO7yE5CXAZx 維尼

=2∫eusin(u+a)du… or choose an alternative:519Please respect copyright.ＰＥＮＡＮＡWxal8I8MHk
Substitute e√x8964 copyright protection515ＰＥＮＡＮＡiQoNh4DWOs 維尼

Now solving:8964 copyright protection515ＰＥＮＡＮＡxs4ZkPCKel 維尼

∫eusin(u+a)du8964 copyright protection515ＰＥＮＡＮＡfRWG9xJXea 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection515ＰＥＮＡＮＡeIMgkx7qbO 維尼

First time:8964 copyright protection515ＰＥＮＡＮＡ0U1RVYvDCi 維尼

f=sin(u+a),g′=eu8964 copyright protection515ＰＥＮＡＮＡ4HQUzVufPA 維尼

↓ steps↓ steps8964 copyright protection515ＰＥＮＡＮＡMckVl9u9qr 維尼

f′=cos(u+a),g=eu:8964 copyright protection515ＰＥＮＡＮＡyVzxpOjZ2T 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection515ＰＥＮＡＮＡycFjlJghoW 維尼

Second time:8964 copyright protection515ＰＥＮＡＮＡju6bq7A8sV 維尼

f=cos(u+a),g′=eu8964 copyright protection515ＰＥＮＡＮＡwK6heFhOch 維尼

↓ steps↓ steps8964 copyright protection515ＰＥＮＡＮＡxwius7qXrm 維尼

f′=−sin(u+a),g=eu:8964 copyright protection515ＰＥＮＡＮＡ1W9UCSsXCf 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection515ＰＥＮＡＮＡCCgmbn2fKB 維尼

Apply linearity:8964 copyright protection515ＰＥＮＡＮＡiTmGFfM6I6 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection515ＰＥＮＡＮＡqpaRjzr9iL 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection515ＰＥＮＡＮＡHdg6eA4zGO 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection515ＰＥＮＡＮＡWUvpwBSpAR 維尼

Plug in solved integrals:8964 copyright protection515ＰＥＮＡＮＡ6cI6MSP8rn 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection515ＰＥＮＡＮＡ4HhcdJnxwQ 維尼

Undo substitution u=√x:8964 copyright protection515ＰＥＮＡＮＡRPhbEPvVig 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection515ＰＥＮＡＮＡLGo3TCMtm2 維尼

The problem is solved:8964 copyright protection515ＰＥＮＡＮＡbQIrRLXgyp 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection515ＰＥＮＡＮＡv7oy6v26K8 維尼

Rewrite/simplify:8964 copyright protection515ＰＥＮＡＮＡYh0gekJ3Bu 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection515ＰＥＮＡＮＡMaPAsXvrKB 維尼

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