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恐懼感8964 copyright protection494ＰＥＮＡＮＡTWTOLgaBAy 維尼

498Please respect copyright.ＰＥＮＡＮＡN5gSMqcZzK
8964 copyright protection494ＰＥＮＡＮＡYpcagzN0os 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection494ＰＥＮＡＮＡllv5bE1EMb 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection494ＰＥＮＡＮＡOhFiDD1pVG 維尼

=2∫eusin(u+a)du… or choose an alternative:498Please respect copyright.ＰＥＮＡＮＡJSj1EEt3Z4
Substitute e√x8964 copyright protection494ＰＥＮＡＮＡUCNFGjVq9L 維尼

Now solving:8964 copyright protection494ＰＥＮＡＮＡ6upXUJ5BG9 維尼

∫eusin(u+a)du8964 copyright protection494ＰＥＮＡＮＡih34aPP72u 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection494ＰＥＮＡＮＡ6OcGecSAVn 維尼

First time:8964 copyright protection494ＰＥＮＡＮＡCDhNac64dP 維尼

f=sin(u+a),g′=eu8964 copyright protection494ＰＥＮＡＮＡF4TiWcJa9w 維尼

↓ steps↓ steps8964 copyright protection494ＰＥＮＡＮＡ8VWrk6eIE6 維尼

f′=cos(u+a),g=eu:8964 copyright protection494ＰＥＮＡＮＡyaEbRBHwQV 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection494ＰＥＮＡＮＡA42Gy6rnEq 維尼

Second time:8964 copyright protection494ＰＥＮＡＮＡ97ZiQTAFM1 維尼

f=cos(u+a),g′=eu8964 copyright protection494ＰＥＮＡＮＡcr7yYxaVtJ 維尼

↓ steps↓ steps8964 copyright protection494ＰＥＮＡＮＡWuP16ilbrN 維尼

f′=−sin(u+a),g=eu:8964 copyright protection494ＰＥＮＡＮＡA1IpdC5uwN 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection494ＰＥＮＡＮＡFJN6YDcm4y 維尼

Apply linearity:8964 copyright protection494ＰＥＮＡＮＡ6sHkgcGT7y 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection494ＰＥＮＡＮＡeI6vF6rwsZ 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection494ＰＥＮＡＮＡcGiXpLC2Ek 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection494ＰＥＮＡＮＡY5r44QKFIk 維尼

Plug in solved integrals:8964 copyright protection494ＰＥＮＡＮＡtJ97mlJCbY 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection494ＰＥＮＡＮＡlR3KLVIAMz 維尼

Undo substitution u=√x:8964 copyright protection494ＰＥＮＡＮＡrJsbdKrhFQ 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection494ＰＥＮＡＮＡqEJg1j9DnR 維尼

The problem is solved:8964 copyright protection494ＰＥＮＡＮＡ3D6ESzOf1A 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection494ＰＥＮＡＮＡ592qasRwIL 維尼

Rewrite/simplify:8964 copyright protection494ＰＥＮＡＮＡ4jLgMoc518 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection494ＰＥＮＡＮＡM68U1ImLeN 維尼

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