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恐懼感8964 copyright protection363ＰＥＮＡＮＡLVPvK9gEeY 維尼

367Please respect copyright.ＰＥＮＡＮＡtXChk5mqqf
8964 copyright protection363ＰＥＮＡＮＡl7vkvxAueS 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection363ＰＥＮＡＮＡcqpUGg5z4X 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection363ＰＥＮＡＮＡNXZdxct7x6 維尼

=2∫eusin(u+a)du… or choose an alternative:367Please respect copyright.ＰＥＮＡＮＡfo7bFNlvqq
Substitute e√x8964 copyright protection363ＰＥＮＡＮＡCZ8JDOUv5n 維尼

Now solving:8964 copyright protection363ＰＥＮＡＮＡC0jkWmLDuv 維尼

∫eusin(u+a)du8964 copyright protection363ＰＥＮＡＮＡIoKhRApBrq 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection363ＰＥＮＡＮＡYckxcUOOs3 維尼

First time:8964 copyright protection363ＰＥＮＡＮＡ4h50MQUkwU 維尼

f=sin(u+a),g′=eu8964 copyright protection363ＰＥＮＡＮＡwDnJzPl43c 維尼

↓ steps↓ steps8964 copyright protection363ＰＥＮＡＮＡjBQBAOaGAS 維尼

f′=cos(u+a),g=eu:8964 copyright protection363ＰＥＮＡＮＡJhNxVFmdgI 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection363ＰＥＮＡＮＡxiJSfgwMpp 維尼

Second time:8964 copyright protection363ＰＥＮＡＮＡWml49WextX 維尼

f=cos(u+a),g′=eu8964 copyright protection363ＰＥＮＡＮＡTSfZLbYgqf 維尼

↓ steps↓ steps8964 copyright protection363ＰＥＮＡＮＡ0vfNkeZ0rK 維尼

f′=−sin(u+a),g=eu:8964 copyright protection363ＰＥＮＡＮＡzmCrc1SAOd 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection363ＰＥＮＡＮＡZwikUhd7Sd 維尼

Apply linearity:8964 copyright protection363ＰＥＮＡＮＡ1k73OO7G4m 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection363ＰＥＮＡＮＡAlB1edTVTJ 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection363ＰＥＮＡＮＡrhr0YsJANg 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection363ＰＥＮＡＮＡITMI0am2fO 維尼

Plug in solved integrals:8964 copyright protection363ＰＥＮＡＮＡNVWmV1KESy 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection363ＰＥＮＡＮＡURrVIG98zl 維尼

Undo substitution u=√x:8964 copyright protection363ＰＥＮＡＮＡMypPBHJZyj 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection363ＰＥＮＡＮＡCTwsRTWSIz 維尼

The problem is solved:8964 copyright protection363ＰＥＮＡＮＡlQiyYYKvSZ 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection363ＰＥＮＡＮＡ3y0pPCU2Bi 維尼

Rewrite/simplify:8964 copyright protection363ＰＥＮＡＮＡQL7CV8iau6 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection363ＰＥＮＡＮＡ13GFHsXblc 維尼

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