x

恐懼感8964 copyright protection453ＰＥＮＡＮＡPEDZkWDmFw 維尼

457Please respect copyright.ＰＥＮＡＮＡrKXDyNJWIF
8964 copyright protection453ＰＥＮＡＮＡ8CV60kRfS3 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection453ＰＥＮＡＮＡ1nRdGYtLId 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection453ＰＥＮＡＮＡ0LsZMj5dqy 維尼

=2∫eusin(u+a)du… or choose an alternative:457Please respect copyright.ＰＥＮＡＮＡ8pjEUhDhBe
Substitute e√x8964 copyright protection453ＰＥＮＡＮＡeAB52c32AY 維尼

Now solving:8964 copyright protection453ＰＥＮＡＮＡtzMc3N1CeW 維尼

∫eusin(u+a)du8964 copyright protection453ＰＥＮＡＮＡjpJ8EDojsB 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection453ＰＥＮＡＮＡRBN3h46sOS 維尼

First time:8964 copyright protection453ＰＥＮＡＮＡ1g58eZYzVD 維尼

f=sin(u+a),g′=eu8964 copyright protection453ＰＥＮＡＮＡCp6S1h6kgO 維尼

↓ steps↓ steps8964 copyright protection453ＰＥＮＡＮＡAcC3LmeTh9 維尼

f′=cos(u+a),g=eu:8964 copyright protection453ＰＥＮＡＮＡbLel87INYP 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection453ＰＥＮＡＮＡV5hdwiX2Yo 維尼

Second time:8964 copyright protection453ＰＥＮＡＮＡAYUPdCH2py 維尼

f=cos(u+a),g′=eu8964 copyright protection453ＰＥＮＡＮＡUA01373qoj 維尼

↓ steps↓ steps8964 copyright protection453ＰＥＮＡＮＡ155sBLb2nS 維尼

f′=−sin(u+a),g=eu:8964 copyright protection453ＰＥＮＡＮＡ6MEr4s5TBT 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection453ＰＥＮＡＮＡCbOmCe9ZG2 維尼

Apply linearity:8964 copyright protection453ＰＥＮＡＮＡ0SXAJohDRK 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection453ＰＥＮＡＮＡEj4TtPlQGW 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection453ＰＥＮＡＮＡcX659NQuDd 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection453ＰＥＮＡＮＡitdElzfq0P 維尼

Plug in solved integrals:8964 copyright protection453ＰＥＮＡＮＡAPjD4dwIBo 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection453ＰＥＮＡＮＡI5g9gBEhkW 維尼

Undo substitution u=√x:8964 copyright protection453ＰＥＮＡＮＡAuhSXzkXNn 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection453ＰＥＮＡＮＡVOZ8ep5v8a 維尼

The problem is solved:8964 copyright protection453ＰＥＮＡＮＡQbfTxjoURb 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection453ＰＥＮＡＮＡ7lGIpgr34b 維尼

Rewrite/simplify:8964 copyright protection453ＰＥＮＡＮＡR4oTEZupOI 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection453ＰＥＮＡＮＡvmxG0xblEy 維尼

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