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恐懼感8964 copyright protection549ＰＥＮＡＮＡ9c4nP5lMF4 維尼

553Please respect copyright.ＰＥＮＡＮＡRy3SvY42al
8964 copyright protection549ＰＥＮＡＮＡRNolPfMQCv 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection549ＰＥＮＡＮＡhXl8L5fJwp 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection549ＰＥＮＡＮＡj8dGtnrjWu 維尼

=2∫eusin(u+a)du… or choose an alternative:553Please respect copyright.ＰＥＮＡＮＡtYIA1vJ4LL
Substitute e√x8964 copyright protection549ＰＥＮＡＮＡaKY1gFZ8Xc 維尼

Now solving:8964 copyright protection549ＰＥＮＡＮＡRrcWTd3BYo 維尼

∫eusin(u+a)du8964 copyright protection549ＰＥＮＡＮＡuqxZfToaUy 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection549ＰＥＮＡＮＡNXmU4Bj9gZ 維尼

First time:8964 copyright protection549ＰＥＮＡＮＡ6BdvLaLklL 維尼

f=sin(u+a),g′=eu8964 copyright protection549ＰＥＮＡＮＡdZ0K7vAf0e 維尼

↓ steps↓ steps8964 copyright protection549ＰＥＮＡＮＡiA5AsVhUJA 維尼

f′=cos(u+a),g=eu:8964 copyright protection549ＰＥＮＡＮＡYkg23blotR 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection549ＰＥＮＡＮＡaqzJPQupzv 維尼

Second time:8964 copyright protection549ＰＥＮＡＮＡI00xpMm8kg 維尼

f=cos(u+a),g′=eu8964 copyright protection549ＰＥＮＡＮＡi7fn1Kmgem 維尼

↓ steps↓ steps8964 copyright protection549ＰＥＮＡＮＡnmngz1peOK 維尼

f′=−sin(u+a),g=eu:8964 copyright protection549ＰＥＮＡＮＡJqzZAJZ1Gt 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection549ＰＥＮＡＮＡ24VIv62AAz 維尼

Apply linearity:8964 copyright protection549ＰＥＮＡＮＡubQMqBu2Uq 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection549ＰＥＮＡＮＡgoAl3byDRR 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection549ＰＥＮＡＮＡ7jKzpaR8t3 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection549ＰＥＮＡＮＡBB1FtygiHC 維尼

Plug in solved integrals:8964 copyright protection549ＰＥＮＡＮＡZQ54PaDP50 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection549ＰＥＮＡＮＡFF9y6HJiom 維尼

Undo substitution u=√x:8964 copyright protection549ＰＥＮＡＮＡBRMTaj4niE 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection549ＰＥＮＡＮＡ90zlJcMGBC 維尼

The problem is solved:8964 copyright protection549ＰＥＮＡＮＡxyk41DKDTt 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection549ＰＥＮＡＮＡ5Cki5oyMYZ 維尼

Rewrite/simplify:8964 copyright protection549ＰＥＮＡＮＡFRkXwi7mWf 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection549ＰＥＮＡＮＡLcTmSUyY8X 維尼

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