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恐懼感8964 copyright protection495ＰＥＮＡＮＡADYcEjY9q3 維尼

499Please respect copyright.ＰＥＮＡＮＡnCkzs7LZdc
8964 copyright protection495ＰＥＮＡＮＡDAg07qH5TS 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection495ＰＥＮＡＮＡ57cqlabXgD 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection495ＰＥＮＡＮＡYseRngnYCb 維尼

=2∫eusin(u+a)du… or choose an alternative:499Please respect copyright.ＰＥＮＡＮＡNiHrkBErRS
Substitute e√x8964 copyright protection495ＰＥＮＡＮＡtoW3SbFUVt 維尼

Now solving:8964 copyright protection495ＰＥＮＡＮＡXaVKgSb122 維尼

∫eusin(u+a)du8964 copyright protection495ＰＥＮＡＮＡD6UT3GC3Ca 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection495ＰＥＮＡＮＡxbWhN3eNtA 維尼

First time:8964 copyright protection495ＰＥＮＡＮＡimwfOgcbbb 維尼

f=sin(u+a),g′=eu8964 copyright protection495ＰＥＮＡＮＡMWRZ9pBiQe 維尼

↓ steps↓ steps8964 copyright protection495ＰＥＮＡＮＡKfHm6AcXzj 維尼

f′=cos(u+a),g=eu:8964 copyright protection495ＰＥＮＡＮＡESgVMkdXOV 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection495ＰＥＮＡＮＡhvJfb5xxx4 維尼

Second time:8964 copyright protection495ＰＥＮＡＮＡuM6WDvIrGR 維尼

f=cos(u+a),g′=eu8964 copyright protection495ＰＥＮＡＮＡ4L4Vn6XR59 維尼

↓ steps↓ steps8964 copyright protection495ＰＥＮＡＮＡ3ZHroDV7rb 維尼

f′=−sin(u+a),g=eu:8964 copyright protection495ＰＥＮＡＮＡHCkXV9zP1x 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection495ＰＥＮＡＮＡwN8K6H4gZi 維尼

Apply linearity:8964 copyright protection495ＰＥＮＡＮＡJYRrpp9IVt 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection495ＰＥＮＡＮＡDRMDllFdOg 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection495ＰＥＮＡＮＡpU8D0deHdV 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection495ＰＥＮＡＮＡXlcIKl8Czh 維尼

Plug in solved integrals:8964 copyright protection495ＰＥＮＡＮＡb9qgbR1L14 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection495ＰＥＮＡＮＡmkTJWKqY6C 維尼

Undo substitution u=√x:8964 copyright protection495ＰＥＮＡＮＡAn747Fg31c 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection495ＰＥＮＡＮＡyaWh7w6no3 維尼

The problem is solved:8964 copyright protection495ＰＥＮＡＮＡW90yqGZHm2 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection495ＰＥＮＡＮＡp5uxKfHd3k 維尼

Rewrite/simplify:8964 copyright protection495ＰＥＮＡＮＡiaZasa7niT 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection495ＰＥＮＡＮＡCOL5llF25Y 維尼

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