x

恐懼感8964 copyright protection375ＰＥＮＡＮＡiz5OjBtVsL 維尼

379Please respect copyright.ＰＥＮＡＮＡcLZCjausq0
8964 copyright protection375ＰＥＮＡＮＡHocXzventP 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection375ＰＥＮＡＮＡzDh4Xot697 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection375ＰＥＮＡＮＡaDClhQHD5w 維尼

=2∫eusin(u+a)du… or choose an alternative:379Please respect copyright.ＰＥＮＡＮＡ4N7DlfiOdX
Substitute e√x8964 copyright protection375ＰＥＮＡＮＡrmZkB9fL8Q 維尼

Now solving:8964 copyright protection375ＰＥＮＡＮＡasPZHIp93V 維尼

∫eusin(u+a)du8964 copyright protection375ＰＥＮＡＮＡdmW30eXDcZ 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection375ＰＥＮＡＮＡcf6Vb5DMec 維尼

First time:8964 copyright protection375ＰＥＮＡＮＡAsNj46DuVH 維尼

f=sin(u+a),g′=eu8964 copyright protection375ＰＥＮＡＮＡ1wlukVKki3 維尼

↓ steps↓ steps8964 copyright protection375ＰＥＮＡＮＡXv5Fk6nX7M 維尼

f′=cos(u+a),g=eu:8964 copyright protection375ＰＥＮＡＮＡwQKXtEZzim 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection375ＰＥＮＡＮＡfqGqRTafwG 維尼

Second time:8964 copyright protection375ＰＥＮＡＮＡHHlT6Atz4v 維尼

f=cos(u+a),g′=eu8964 copyright protection375ＰＥＮＡＮＡPF0HmAsyxj 維尼

↓ steps↓ steps8964 copyright protection375ＰＥＮＡＮＡdiz478KUwP 維尼

f′=−sin(u+a),g=eu:8964 copyright protection375ＰＥＮＡＮＡddJqsvu3Lg 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection375ＰＥＮＡＮＡJDsSohKdoV 維尼

Apply linearity:8964 copyright protection375ＰＥＮＡＮＡ7SnXWrVyyp 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection375ＰＥＮＡＮＡBjObnPZRWW 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection375ＰＥＮＡＮＡYvomGNBNi5 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection375ＰＥＮＡＮＡEpzbxw5amE 維尼

Plug in solved integrals:8964 copyright protection375ＰＥＮＡＮＡAFGAk0AAUq 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection375ＰＥＮＡＮＡYRpe3yhjUB 維尼

Undo substitution u=√x:8964 copyright protection375ＰＥＮＡＮＡVBbsj6YCZH 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection375ＰＥＮＡＮＡLA6VpmSPh6 維尼

The problem is solved:8964 copyright protection375ＰＥＮＡＮＡsdTWPhNOfL 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection375ＰＥＮＡＮＡTVdcQap0ED 維尼

Rewrite/simplify:8964 copyright protection375ＰＥＮＡＮＡyB40ZtivJU 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection375ＰＥＮＡＮＡDya9QiYcxN 維尼

15.158.61.48