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恐懼感8964 copyright protection413ＰＥＮＡＮＡOFiovhi1HG 維尼

417Please respect copyright.ＰＥＮＡＮＡ9sh4m06ugb
8964 copyright protection413ＰＥＮＡＮＡio2Om8DSpe 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection413ＰＥＮＡＮＡNiWnUzY3D1 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection413ＰＥＮＡＮＡ8ZJcUVlViK 維尼

=2∫eusin(u+a)du… or choose an alternative:417Please respect copyright.ＰＥＮＡＮＡIfDr3J0C3c
Substitute e√x8964 copyright protection413ＰＥＮＡＮＡSwi7gwESYg 維尼

Now solving:8964 copyright protection413ＰＥＮＡＮＡlwmXwwK4h9 維尼

∫eusin(u+a)du8964 copyright protection413ＰＥＮＡＮＡUwh1skbcx1 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection413ＰＥＮＡＮＡusUFSuYzoV 維尼

First time:8964 copyright protection413ＰＥＮＡＮＡw7v2twx4Ih 維尼

f=sin(u+a),g′=eu8964 copyright protection413ＰＥＮＡＮＡkQvHEVl9Zi 維尼

↓ steps↓ steps8964 copyright protection413ＰＥＮＡＮＡK0UuYAna4o 維尼

f′=cos(u+a),g=eu:8964 copyright protection413ＰＥＮＡＮＡGGCGhwNDbd 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection413ＰＥＮＡＮＡS4pzVKyXy5 維尼

Second time:8964 copyright protection413ＰＥＮＡＮＡnEADYo107q 維尼

f=cos(u+a),g′=eu8964 copyright protection413ＰＥＮＡＮＡ04FU2aMmmV 維尼

↓ steps↓ steps8964 copyright protection413ＰＥＮＡＮＡqxcrSVQcJW 維尼

f′=−sin(u+a),g=eu:8964 copyright protection413ＰＥＮＡＮＡhdYU9C6B3J 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection413ＰＥＮＡＮＡJIT32HGMcU 維尼

Apply linearity:8964 copyright protection413ＰＥＮＡＮＡ4oHKMuGhrj 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection413ＰＥＮＡＮＡ9iEcTM4PDl 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection413ＰＥＮＡＮＡRS138UU5nQ 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection413ＰＥＮＡＮＡwH3maQAZUU 維尼

Plug in solved integrals:8964 copyright protection413ＰＥＮＡＮＡOFUopZoVBH 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection413ＰＥＮＡＮＡfKr73a4uqK 維尼

Undo substitution u=√x:8964 copyright protection413ＰＥＮＡＮＡTKqQe5qSvt 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection413ＰＥＮＡＮＡvMbofzMKhq 維尼

The problem is solved:8964 copyright protection413ＰＥＮＡＮＡWhMXCrM6Ke 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection413ＰＥＮＡＮＡcEbi3zvd0v 維尼

Rewrite/simplify:8964 copyright protection413ＰＥＮＡＮＡsM4m8dDg9e 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection413ＰＥＮＡＮＡr670RlLYhQ 維尼

18.118.216.5