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恐懼感8964 copyright protection513ＰＥＮＡＮＡ3IqeYGsSAI 維尼

517Please respect copyright.ＰＥＮＡＮＡSdRW7eXlLa
8964 copyright protection513ＰＥＮＡＮＡCggS9JUfNY 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection513ＰＥＮＡＮＡW154eHBsba 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection513ＰＥＮＡＮＡttJvlNnQ37 維尼

=2∫eusin(u+a)du… or choose an alternative:517Please respect copyright.ＰＥＮＡＮＡbxOCGHnumC
Substitute e√x8964 copyright protection513ＰＥＮＡＮＡ9wnapjZySF 維尼

Now solving:8964 copyright protection513ＰＥＮＡＮＡb8nEs7ZmBM 維尼

∫eusin(u+a)du8964 copyright protection513ＰＥＮＡＮＡy4zEuXGl55 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection513ＰＥＮＡＮＡbXHBqNYp0B 維尼

First time:8964 copyright protection513ＰＥＮＡＮＡINVLKF0KcA 維尼

f=sin(u+a),g′=eu8964 copyright protection513ＰＥＮＡＮＡzArqhbmz6M 維尼

↓ steps↓ steps8964 copyright protection513ＰＥＮＡＮＡXOaedLFYZi 維尼

f′=cos(u+a),g=eu:8964 copyright protection513ＰＥＮＡＮＡdv15wS45KJ 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection513ＰＥＮＡＮＡQj486SPDxy 維尼

Second time:8964 copyright protection513ＰＥＮＡＮＡOUkLLYvIfr 維尼

f=cos(u+a),g′=eu8964 copyright protection513ＰＥＮＡＮＡpXOk02TGeF 維尼

↓ steps↓ steps8964 copyright protection513ＰＥＮＡＮＡHONlGSKo8j 維尼

f′=−sin(u+a),g=eu:8964 copyright protection513ＰＥＮＡＮＡplzxT08VXp 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection513ＰＥＮＡＮＡaR0HQApafA 維尼

Apply linearity:8964 copyright protection513ＰＥＮＡＮＡJvzbSb93Rs 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection513ＰＥＮＡＮＡ8otTfk1boq 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection513ＰＥＮＡＮＡJdG7adEvKz 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection513ＰＥＮＡＮＡae4Gtbm2SH 維尼

Plug in solved integrals:8964 copyright protection513ＰＥＮＡＮＡTuzzQbzJVb 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection513ＰＥＮＡＮＡLnFmiQyLsy 維尼

Undo substitution u=√x:8964 copyright protection513ＰＥＮＡＮＡtCiydwviSR 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection513ＰＥＮＡＮＡcYr0hYreRe 維尼

The problem is solved:8964 copyright protection513ＰＥＮＡＮＡqxj3xzu0Nv 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection513ＰＥＮＡＮＡMWvLi1DkgY 維尼

Rewrite/simplify:8964 copyright protection513ＰＥＮＡＮＡeZtvq1SXqJ 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection513ＰＥＮＡＮＡaNPTD4qP44 維尼

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