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恐懼感8964 copyright protection490ＰＥＮＡＮＡ9fssENd0bN 維尼

494Please respect copyright.ＰＥＮＡＮＡ6d3udGBhoW
8964 copyright protection490ＰＥＮＡＮＡ7F70rnB8cw 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection490ＰＥＮＡＮＡsCet326etQ 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection490ＰＥＮＡＮＡSN4uyVISUf 維尼

=2∫eusin(u+a)du… or choose an alternative:494Please respect copyright.ＰＥＮＡＮＡI4xVXSHXIi
Substitute e√x8964 copyright protection490ＰＥＮＡＮＡ06y4IfENGm 維尼

Now solving:8964 copyright protection490ＰＥＮＡＮＡhaVSFRCtIV 維尼

∫eusin(u+a)du8964 copyright protection490ＰＥＮＡＮＡi6aNkEqjE5 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection490ＰＥＮＡＮＡtm55aaiE02 維尼

First time:8964 copyright protection490ＰＥＮＡＮＡ3DhFVBCcj2 維尼

f=sin(u+a),g′=eu8964 copyright protection490ＰＥＮＡＮＡqsZ5Cxyh11 維尼

↓ steps↓ steps8964 copyright protection490ＰＥＮＡＮＡ1tonJ7VZ7g 維尼

f′=cos(u+a),g=eu:8964 copyright protection490ＰＥＮＡＮＡi2KW6mWY6B 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection490ＰＥＮＡＮＡkAUfJJ6VQJ 維尼

Second time:8964 copyright protection490ＰＥＮＡＮＡdIvLKI8a6T 維尼

f=cos(u+a),g′=eu8964 copyright protection490ＰＥＮＡＮＡSkKktTB5r9 維尼

↓ steps↓ steps8964 copyright protection490ＰＥＮＡＮＡZIkH8GFmuZ 維尼

f′=−sin(u+a),g=eu:8964 copyright protection490ＰＥＮＡＮＡzN97UfDAaU 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection490ＰＥＮＡＮＡXLpUO0LgkZ 維尼

Apply linearity:8964 copyright protection490ＰＥＮＡＮＡMOUPiX4xCw 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection490ＰＥＮＡＮＡQmmBplzC6c 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection490ＰＥＮＡＮＡjAVEXJf3oB 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection490ＰＥＮＡＮＡHDb8BHFcql 維尼

Plug in solved integrals:8964 copyright protection490ＰＥＮＡＮＡKdGIs8UdpN 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection490ＰＥＮＡＮＡ2PYGoanh4U 維尼

Undo substitution u=√x:8964 copyright protection490ＰＥＮＡＮＡf8nLuleZss 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection490ＰＥＮＡＮＡY01vb87CQi 維尼

The problem is solved:8964 copyright protection490ＰＥＮＡＮＡzice3HLddv 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection490ＰＥＮＡＮＡrb9m2miTun 維尼

Rewrite/simplify:8964 copyright protection490ＰＥＮＡＮＡkIDqH9j1fp 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection490ＰＥＮＡＮＡkouSpq8k8b 維尼

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