恐懼感8964 copyright protection562ＰＥＮＡＮＡpZdKNCQcel 維尼

566Please respect copyright.ＰＥＮＡＮＡi36Ob6gmOE
8964 copyright protection562ＰＥＮＡＮＡKl5Vh4IzB5 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection562ＰＥＮＡＮＡjYGmeSPCXE 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection562ＰＥＮＡＮＡYnFU7zVU5d 維尼

=2∫eusin(u+a)du… or choose an alternative:566Please respect copyright.ＰＥＮＡＮＡ1xNjz9gu1b
Substitute e√x8964 copyright protection562ＰＥＮＡＮＡpbystqJKra 維尼

Now solving:8964 copyright protection562ＰＥＮＡＮＡJ7ndLLYNfx 維尼

∫eusin(u+a)du8964 copyright protection562ＰＥＮＡＮＡAplgpFyWLB 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection562ＰＥＮＡＮＡkiWHc47RSk 維尼

First time:8964 copyright protection562ＰＥＮＡＮＡKaBEEmrJ9E 維尼

f=sin(u+a),g′=eu8964 copyright protection562ＰＥＮＡＮＡyfD67im9M0 維尼

↓ steps↓ steps8964 copyright protection562ＰＥＮＡＮＡyL1ZaSh02u 維尼

f′=cos(u+a),g=eu:8964 copyright protection562ＰＥＮＡＮＡwnrt0thpZR 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection562ＰＥＮＡＮＡU5bNYJF2Xf 維尼

Second time:8964 copyright protection562ＰＥＮＡＮＡNNuvDrGpLG 維尼

f=cos(u+a),g′=eu8964 copyright protection562ＰＥＮＡＮＡunfy8FVjsR 維尼

↓ steps↓ steps8964 copyright protection562ＰＥＮＡＮＡLjLnUTVULv 維尼

f′=−sin(u+a),g=eu:8964 copyright protection562ＰＥＮＡＮＡCWGzQ77Msu 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection562ＰＥＮＡＮＡTLZ2uB78PY 維尼

Apply linearity:8964 copyright protection562ＰＥＮＡＮＡGgclpFsmFN 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection562ＰＥＮＡＮＡosm2DjuduD 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection562ＰＥＮＡＮＡWAU8BaEoXa 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection562ＰＥＮＡＮＡFQAZWYMzJu 維尼

Plug in solved integrals:8964 copyright protection562ＰＥＮＡＮＡXNzZAHImwh 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection562ＰＥＮＡＮＡTcyeaTNFkt 維尼

Undo substitution u=√x:8964 copyright protection562ＰＥＮＡＮＡJjKQTQXtRu 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection562ＰＥＮＡＮＡqIYaaYw8Pm 維尼

The problem is solved:8964 copyright protection562ＰＥＮＡＮＡKA4KSXEywv 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection562ＰＥＮＡＮＡyaM8Cr5Yaf 維尼

Rewrite/simplify:8964 copyright protection562ＰＥＮＡＮＡyNnhsWHco8 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection562ＰＥＮＡＮＡ0xkbUyXKTV 維尼

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