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恐懼感8964 copyright protection374ＰＥＮＡＮＡdziMWRLioN 維尼

378Please respect copyright.ＰＥＮＡＮＡpTPoDdcakF
8964 copyright protection374ＰＥＮＡＮＡjbDmHLNK6Y 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection374ＰＥＮＡＮＡcM0rKA5Sge 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection374ＰＥＮＡＮＡYiM086dPAE 維尼

=2∫eusin(u+a)du… or choose an alternative:378Please respect copyright.ＰＥＮＡＮＡw5KrIfYqtJ
Substitute e√x8964 copyright protection374ＰＥＮＡＮＡF9rovlOyEz 維尼

Now solving:8964 copyright protection374ＰＥＮＡＮＡgyEL0VI0WU 維尼

∫eusin(u+a)du8964 copyright protection374ＰＥＮＡＮＡLQp4AkkYKv 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection374ＰＥＮＡＮＡuAMz5o9j4n 維尼

First time:8964 copyright protection374ＰＥＮＡＮＡFhpOVUvQs9 維尼

f=sin(u+a),g′=eu8964 copyright protection374ＰＥＮＡＮＡXlKaKBvQAn 維尼

↓ steps↓ steps8964 copyright protection374ＰＥＮＡＮＡWE7DwKAUDw 維尼

f′=cos(u+a),g=eu:8964 copyright protection374ＰＥＮＡＮＡKkMGpSNSVS 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection374ＰＥＮＡＮＡKWWOaJNGDE 維尼

Second time:8964 copyright protection374ＰＥＮＡＮＡRqviUidTQI 維尼

f=cos(u+a),g′=eu8964 copyright protection374ＰＥＮＡＮＡdVkDqaLHWu 維尼

↓ steps↓ steps8964 copyright protection374ＰＥＮＡＮＡKPKm0GPp2E 維尼

f′=−sin(u+a),g=eu:8964 copyright protection374ＰＥＮＡＮＡyytNGtDLZA 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection374ＰＥＮＡＮＡsyvNe1ARPv 維尼

Apply linearity:8964 copyright protection374ＰＥＮＡＮＡwFjjUE77hb 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection374ＰＥＮＡＮＡW6nzw440s6 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection374ＰＥＮＡＮＡrIkbP9zbCq 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection374ＰＥＮＡＮＡ3YOJ1lnvsT 維尼

Plug in solved integrals:8964 copyright protection374ＰＥＮＡＮＡBgHIDKpU6O 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection374ＰＥＮＡＮＡQfFOD6XzAS 維尼

Undo substitution u=√x:8964 copyright protection374ＰＥＮＡＮＡLOo930uVHk 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection374ＰＥＮＡＮＡx5qKIpYnGb 維尼

The problem is solved:8964 copyright protection374ＰＥＮＡＮＡpuOHK1sJY0 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection374ＰＥＮＡＮＡGsd8FTaICa 維尼

Rewrite/simplify:8964 copyright protection374ＰＥＮＡＮＡDDE00jQION 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection374ＰＥＮＡＮＡ6OUB6wwMGG 維尼

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