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恐懼感8964 copyright protection440ＰＥＮＡＮＡSmIK8NNQjK 維尼

444Please respect copyright.ＰＥＮＡＮＡobe54Wb0B4
8964 copyright protection440ＰＥＮＡＮＡJDZCV2iuad 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection440ＰＥＮＡＮＡMbfXnDKs3j 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection440ＰＥＮＡＮＡPXDIqfIkBX 維尼

=2∫eusin(u+a)du… or choose an alternative:444Please respect copyright.ＰＥＮＡＮＡ1odJ37cIIA
Substitute e√x8964 copyright protection440ＰＥＮＡＮＡLwL37QhXao 維尼

Now solving:8964 copyright protection440ＰＥＮＡＮＡfEzjX0FhJM 維尼

∫eusin(u+a)du8964 copyright protection440ＰＥＮＡＮＡ3bZeFVgn5P 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection440ＰＥＮＡＮＡjbOiKVAagE 維尼

First time:8964 copyright protection440ＰＥＮＡＮＡ4nVFJDdMFK 維尼

f=sin(u+a),g′=eu8964 copyright protection440ＰＥＮＡＮＡcBUNK6Hnt6 維尼

↓ steps↓ steps8964 copyright protection440ＰＥＮＡＮＡgNH5Zf5hB3 維尼

f′=cos(u+a),g=eu:8964 copyright protection440ＰＥＮＡＮＡjomUZ3ecsB 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection440ＰＥＮＡＮＡdQ7PBPswKY 維尼

Second time:8964 copyright protection440ＰＥＮＡＮＡYRcw4qQK0f 維尼

f=cos(u+a),g′=eu8964 copyright protection440ＰＥＮＡＮＡ7UZz87oxZd 維尼

↓ steps↓ steps8964 copyright protection440ＰＥＮＡＮＡOGVTWhQamt 維尼

f′=−sin(u+a),g=eu:8964 copyright protection440ＰＥＮＡＮＡ1GWtQk8pni 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection440ＰＥＮＡＮＡSp66cxkLaU 維尼

Apply linearity:8964 copyright protection440ＰＥＮＡＮＡuVs502Xc6V 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection440ＰＥＮＡＮＡaHRzWDh9gX 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection440ＰＥＮＡＮＡSNLaUwOSFS 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection440ＰＥＮＡＮＡ2jbwHfFvgK 維尼

Plug in solved integrals:8964 copyright protection440ＰＥＮＡＮＡ85yve3s6qF 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection440ＰＥＮＡＮＡLGH4VODC4k 維尼

Undo substitution u=√x:8964 copyright protection440ＰＥＮＡＮＡFHBSeQKjb5 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection440ＰＥＮＡＮＡ6C2PTVOWxV 維尼

The problem is solved:8964 copyright protection440ＰＥＮＡＮＡFdUnEqYvvJ 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection440ＰＥＮＡＮＡlYXjXaziWH 維尼

Rewrite/simplify:8964 copyright protection440ＰＥＮＡＮＡG92MWJ02ri 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection440ＰＥＮＡＮＡrgyySj2df1 維尼

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