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恐懼感8964 copyright protection534ＰＥＮＡＮＡXeqY0vREWX 維尼

538Please respect copyright.ＰＥＮＡＮＡmVX5NKPyeU
8964 copyright protection534ＰＥＮＡＮＡqvf80ynH1r 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection534ＰＥＮＡＮＡlbQ0cc1Itt 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection534ＰＥＮＡＮＡkDcbPbJfxR 維尼

=2∫eusin(u+a)du… or choose an alternative:538Please respect copyright.ＰＥＮＡＮＡ6CoBTvYhpg
Substitute e√x8964 copyright protection534ＰＥＮＡＮＡ8gqoeqngBQ 維尼

Now solving:8964 copyright protection534ＰＥＮＡＮＡzY9df2D9rz 維尼

∫eusin(u+a)du8964 copyright protection534ＰＥＮＡＮＡfZksn9Xu5M 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection534ＰＥＮＡＮＡRNExtI2kam 維尼

First time:8964 copyright protection534ＰＥＮＡＮＡGrkMC56wTP 維尼

f=sin(u+a),g′=eu8964 copyright protection534ＰＥＮＡＮＡvq54d4Kd6I 維尼

↓ steps↓ steps8964 copyright protection534ＰＥＮＡＮＡHeGFEryhmY 維尼

f′=cos(u+a),g=eu:8964 copyright protection534ＰＥＮＡＮＡfji2md2mMZ 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection534ＰＥＮＡＮＡL5GaYJtmBu 維尼

Second time:8964 copyright protection534ＰＥＮＡＮＡbhAyjt8crJ 維尼

f=cos(u+a),g′=eu8964 copyright protection534ＰＥＮＡＮＡJKBwI374au 維尼

↓ steps↓ steps8964 copyright protection534ＰＥＮＡＮＡQpoYvJQ867 維尼

f′=−sin(u+a),g=eu:8964 copyright protection534ＰＥＮＡＮＡx4mS9hyMdB 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection534ＰＥＮＡＮＡOHkelMqUzy 維尼

Apply linearity:8964 copyright protection534ＰＥＮＡＮＡAX5xPFKruk 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection534ＰＥＮＡＮＡ0JujN2vllN 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection534ＰＥＮＡＮＡUlUhh0Nmbt 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection534ＰＥＮＡＮＡYJXn7o8sqb 維尼

Plug in solved integrals:8964 copyright protection534ＰＥＮＡＮＡnmRzMYvoVN 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection534ＰＥＮＡＮＡHvo79YCVKw 維尼

Undo substitution u=√x:8964 copyright protection534ＰＥＮＡＮＡdOUa1AjK1I 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection534ＰＥＮＡＮＡzJqTn1lh1N 維尼

The problem is solved:8964 copyright protection534ＰＥＮＡＮＡz9HOdT12zs 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection534ＰＥＮＡＮＡfJoFnQS8mF 維尼

Rewrite/simplify:8964 copyright protection534ＰＥＮＡＮＡ0mDLPpE6R2 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection534ＰＥＮＡＮＡ26IwQ4y3hD 維尼

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