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恐懼感8964 copyright protection488ＰＥＮＡＮＡCmiWXuduMi 維尼

492Please respect copyright.ＰＥＮＡＮＡDvWWlfDwZZ
8964 copyright protection488ＰＥＮＡＮＡulIZ24pDDr 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection488ＰＥＮＡＮＡi8SUW0kfM2 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection488ＰＥＮＡＮＡYqj3HYna7M 維尼

=2∫eusin(u+a)du… or choose an alternative:492Please respect copyright.ＰＥＮＡＮＡR1Zjsf4bac
Substitute e√x8964 copyright protection488ＰＥＮＡＮＡES5JYmCcEA 維尼

Now solving:8964 copyright protection488ＰＥＮＡＮＡAfv2HqFrg3 維尼

∫eusin(u+a)du8964 copyright protection488ＰＥＮＡＮＡvL6ob0KGDJ 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection488ＰＥＮＡＮＡQ0BxCdgJ5I 維尼

First time:8964 copyright protection488ＰＥＮＡＮＡCETY3M8oCt 維尼

f=sin(u+a),g′=eu8964 copyright protection488ＰＥＮＡＮＡtgkCVqJfZr 維尼

↓ steps↓ steps8964 copyright protection488ＰＥＮＡＮＡ6sUu7PChGA 維尼

f′=cos(u+a),g=eu:8964 copyright protection488ＰＥＮＡＮＡ1gNDEbbn9b 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection488ＰＥＮＡＮＡPeo2RWGWXA 維尼

Second time:8964 copyright protection488ＰＥＮＡＮＡOmVn7nf2u1 維尼

f=cos(u+a),g′=eu8964 copyright protection488ＰＥＮＡＮＡUxMeIV3kPC 維尼

↓ steps↓ steps8964 copyright protection488ＰＥＮＡＮＡhFeXK1hQW2 維尼

f′=−sin(u+a),g=eu:8964 copyright protection488ＰＥＮＡＮＡXGibt2UAQL 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection488ＰＥＮＡＮＡ9yTuT3J1Gq 維尼

Apply linearity:8964 copyright protection488ＰＥＮＡＮＡijeQbm8phH 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection488ＰＥＮＡＮＡaD4MAeMThO 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection488ＰＥＮＡＮＡ4uDmIsMHt9 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection488ＰＥＮＡＮＡntsBZtTnbN 維尼

Plug in solved integrals:8964 copyright protection488ＰＥＮＡＮＡJVNtiRgrJN 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection488ＰＥＮＡＮＡDHVDhvQcT7 維尼

Undo substitution u=√x:8964 copyright protection488ＰＥＮＡＮＡ3Nw1ltO5q6 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection488ＰＥＮＡＮＡkJiYOcBqMK 維尼

The problem is solved:8964 copyright protection488ＰＥＮＡＮＡqJ5Z1lDI1v 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection488ＰＥＮＡＮＡs2iWhjvFaH 維尼

Rewrite/simplify:8964 copyright protection488ＰＥＮＡＮＡ4bGTftrQDb 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection488ＰＥＮＡＮＡASAvxzLnza 維尼

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