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恐懼感8964 copyright protection528ＰＥＮＡＮＡtXRhf3kcds 維尼

532Please respect copyright.ＰＥＮＡＮＡGXdSAWqRMX
8964 copyright protection528ＰＥＮＡＮＡ3yf0VUGnAF 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection528ＰＥＮＡＮＡwLT6djOc42 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection528ＰＥＮＡＮＡ1yOP5lSICK 維尼

=2∫eusin(u+a)du… or choose an alternative:532Please respect copyright.ＰＥＮＡＮＡJ55YiSBmXK
Substitute e√x8964 copyright protection528ＰＥＮＡＮＡR4RNLtqyP3 維尼

Now solving:8964 copyright protection528ＰＥＮＡＮＡcBYrEQRhco 維尼

∫eusin(u+a)du8964 copyright protection528ＰＥＮＡＮＡ9lZBt0EuER 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection528ＰＥＮＡＮＡYpYbQP8fyH 維尼

First time:8964 copyright protection528ＰＥＮＡＮＡ2ey0Igrfpi 維尼

f=sin(u+a),g′=eu8964 copyright protection528ＰＥＮＡＮＡoLkm8wOUlw 維尼

↓ steps↓ steps8964 copyright protection528ＰＥＮＡＮＡHMQ1dyXaAy 維尼

f′=cos(u+a),g=eu:8964 copyright protection528ＰＥＮＡＮＡnUsJyuJL87 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection528ＰＥＮＡＮＡj27TRYXRz4 維尼

Second time:8964 copyright protection528ＰＥＮＡＮＡLi89PjforH 維尼

f=cos(u+a),g′=eu8964 copyright protection528ＰＥＮＡＮＡlstiZeH3io 維尼

↓ steps↓ steps8964 copyright protection528ＰＥＮＡＮＡ0QeMo6SWeJ 維尼

f′=−sin(u+a),g=eu:8964 copyright protection528ＰＥＮＡＮＡN3uJjljoV9 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection528ＰＥＮＡＮＡCatdSyjQcX 維尼

Apply linearity:8964 copyright protection528ＰＥＮＡＮＡppIdtQJeeZ 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection528ＰＥＮＡＮＡXZJTCCS28W 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection528ＰＥＮＡＮＡyoTaNLMeE2 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection528ＰＥＮＡＮＡv8FtTE585z 維尼

Plug in solved integrals:8964 copyright protection528ＰＥＮＡＮＡ7stnN88yRu 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection528ＰＥＮＡＮＡ0Ilo4LWlM5 維尼

Undo substitution u=√x:8964 copyright protection528ＰＥＮＡＮＡbZTiKkoZLR 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection528ＰＥＮＡＮＡAwgZp8g728 維尼

The problem is solved:8964 copyright protection528ＰＥＮＡＮＡ0RGmoqWtLw 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection528ＰＥＮＡＮＡonWBSvOtXd 維尼

Rewrite/simplify:8964 copyright protection528ＰＥＮＡＮＡnEaRC2tFBC 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection528ＰＥＮＡＮＡ4GTMT1FSUN 維尼

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