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恐懼感8964 copyright protection438ＰＥＮＡＮＡYCKIFsJvsB 維尼

442Please respect copyright.ＰＥＮＡＮＡyZfiqJZu4D
8964 copyright protection438ＰＥＮＡＮＡo1GkZs0wPm 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection438ＰＥＮＡＮＡ9i2vBpfUdR 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection438ＰＥＮＡＮＡUD5yzUazAG 維尼

=2∫eusin(u+a)du… or choose an alternative:442Please respect copyright.ＰＥＮＡＮＡLG927h1LeN
Substitute e√x8964 copyright protection438ＰＥＮＡＮＡPFWfNQgyb7 維尼

Now solving:8964 copyright protection438ＰＥＮＡＮＡ62pwVsgSdZ 維尼

∫eusin(u+a)du8964 copyright protection438ＰＥＮＡＮＡlKlGl13S6w 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection438ＰＥＮＡＮＡUakpl322IL 維尼

First time:8964 copyright protection438ＰＥＮＡＮＡTAHeWTMeCq 維尼

f=sin(u+a),g′=eu8964 copyright protection438ＰＥＮＡＮＡPaaHIFhWVR 維尼

↓ steps↓ steps8964 copyright protection438ＰＥＮＡＮＡVp8RVhgkED 維尼

f′=cos(u+a),g=eu:8964 copyright protection438ＰＥＮＡＮＡCahjmJOeqx 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection438ＰＥＮＡＮＡL2SeqNBRFP 維尼

Second time:8964 copyright protection438ＰＥＮＡＮＡVVSIQTzb7y 維尼

f=cos(u+a),g′=eu8964 copyright protection438ＰＥＮＡＮＡ61IzN1YS8X 維尼

↓ steps↓ steps8964 copyright protection438ＰＥＮＡＮＡVgFatVSH7g 維尼

f′=−sin(u+a),g=eu:8964 copyright protection438ＰＥＮＡＮＡdD3hv7nwuy 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection438ＰＥＮＡＮＡxsyUzzmcix 維尼

Apply linearity:8964 copyright protection438ＰＥＮＡＮＡS0JVP56NJj 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection438ＰＥＮＡＮＡiQApU7eIMl 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection438ＰＥＮＡＮＡ3zLJRFVyK3 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection438ＰＥＮＡＮＡCwGiQjeQop 維尼

Plug in solved integrals:8964 copyright protection438ＰＥＮＡＮＡCjLJxjwNnB 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection438ＰＥＮＡＮＡddtxMlecrt 維尼

Undo substitution u=√x:8964 copyright protection438ＰＥＮＡＮＡdzEcjfZc6E 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection438ＰＥＮＡＮＡrNymmOuuvm 維尼

The problem is solved:8964 copyright protection438ＰＥＮＡＮＡiuSKa0Sy0Z 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection438ＰＥＮＡＮＡL98ZTNILRh 維尼

Rewrite/simplify:8964 copyright protection438ＰＥＮＡＮＡS6CkxrqLg8 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection438ＰＥＮＡＮＡUY02HYj6CJ 維尼

3.16.124.181