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恐懼感8964 copyright protection381ＰＥＮＡＮＡK5QkfY0d9B 維尼

385Please respect copyright.ＰＥＮＡＮＡoleMMnXMaI
8964 copyright protection381ＰＥＮＡＮＡfoEzhD6xja 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection381ＰＥＮＡＮＡdwAAALcjUr 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection381ＰＥＮＡＮＡDTjxlYNeti 維尼

=2∫eusin(u+a)du… or choose an alternative:385Please respect copyright.ＰＥＮＡＮＡuJlnAaLXWm
Substitute e√x8964 copyright protection381ＰＥＮＡＮＡsgALpKlLS0 維尼

Now solving:8964 copyright protection381ＰＥＮＡＮＡAVEoyByZHV 維尼

∫eusin(u+a)du8964 copyright protection381ＰＥＮＡＮＡCXLH8KL3Fb 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection381ＰＥＮＡＮＡcaBffDkaGR 維尼

First time:8964 copyright protection381ＰＥＮＡＮＡihJDFYIb2A 維尼

f=sin(u+a),g′=eu8964 copyright protection381ＰＥＮＡＮＡFd0bqpp6Qj 維尼

↓ steps↓ steps8964 copyright protection381ＰＥＮＡＮＡzu34nJpi3h 維尼

f′=cos(u+a),g=eu:8964 copyright protection381ＰＥＮＡＮＡCszkmT8hln 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection381ＰＥＮＡＮＡbdiI8OVd7L 維尼

Second time:8964 copyright protection381ＰＥＮＡＮＡKWdPTLFrfa 維尼

f=cos(u+a),g′=eu8964 copyright protection381ＰＥＮＡＮＡsYxfdKnRSX 維尼

↓ steps↓ steps8964 copyright protection381ＰＥＮＡＮＡianyy8K7FE 維尼

f′=−sin(u+a),g=eu:8964 copyright protection381ＰＥＮＡＮＡpgG5RULMP0 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection381ＰＥＮＡＮＡbStuLyzwH4 維尼

Apply linearity:8964 copyright protection381ＰＥＮＡＮＡR34fxVf8BV 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection381ＰＥＮＡＮＡsQd46uFVs2 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection381ＰＥＮＡＮＡA7ToN7Xmv4 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection381ＰＥＮＡＮＡLjdqfRzO08 維尼

Plug in solved integrals:8964 copyright protection381ＰＥＮＡＮＡOZGUFqSDAV 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection381ＰＥＮＡＮＡzINUJfZ6IH 維尼

Undo substitution u=√x:8964 copyright protection381ＰＥＮＡＮＡBbaAdJhy9x 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection381ＰＥＮＡＮＡ5o9yyrzS5Z 維尼

The problem is solved:8964 copyright protection381ＰＥＮＡＮＡtM67y4zwKb 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection381ＰＥＮＡＮＡ70RLFBnOHB 維尼

Rewrite/simplify:8964 copyright protection381ＰＥＮＡＮＡnjCkCOhkUg 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection381ＰＥＮＡＮＡ387Y2FErA2 維尼

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