x

恐懼感8964 copyright protection386ＰＥＮＡＮＡ0EZDfRAcmZ 維尼

390Please respect copyright.ＰＥＮＡＮＡQM485xDDzB
8964 copyright protection386ＰＥＮＡＮＡKUtIxUv140 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection386ＰＥＮＡＮＡ2RTts61twb 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection386ＰＥＮＡＮＡPRKb1TKOVJ 維尼

=2∫eusin(u+a)du… or choose an alternative:390Please respect copyright.ＰＥＮＡＮＡNiUvk5u5q0
Substitute e√x8964 copyright protection386ＰＥＮＡＮＡ3B8XyjsZPa 維尼

Now solving:8964 copyright protection386ＰＥＮＡＮＡ8lSR07gcrD 維尼

∫eusin(u+a)du8964 copyright protection386ＰＥＮＡＮＡJpRecgGLgj 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection386ＰＥＮＡＮＡUiTrq9LVGO 維尼

First time:8964 copyright protection386ＰＥＮＡＮＡwhSodBJ1mV 維尼

f=sin(u+a),g′=eu8964 copyright protection386ＰＥＮＡＮＡe5uR7hMxsE 維尼

↓ steps↓ steps8964 copyright protection386ＰＥＮＡＮＡhoGhrHuVQb 維尼

f′=cos(u+a),g=eu:8964 copyright protection386ＰＥＮＡＮＡVommp5BhWU 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection386ＰＥＮＡＮＡbjET2KUdzO 維尼

Second time:8964 copyright protection386ＰＥＮＡＮＡEajEGgarNw 維尼

f=cos(u+a),g′=eu8964 copyright protection386ＰＥＮＡＮＡBRIWqLYZ00 維尼

↓ steps↓ steps8964 copyright protection386ＰＥＮＡＮＡirZsYMaLOh 維尼

f′=−sin(u+a),g=eu:8964 copyright protection386ＰＥＮＡＮＡIm5TYidjHN 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection386ＰＥＮＡＮＡM1rEup8Rj6 維尼

Apply linearity:8964 copyright protection386ＰＥＮＡＮＡ9GhesVrggl 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection386ＰＥＮＡＮＡpXBXiR5S8S 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection386ＰＥＮＡＮＡnWHaAcsqNP 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection386ＰＥＮＡＮＡApGFUr09oq 維尼

Plug in solved integrals:8964 copyright protection386ＰＥＮＡＮＡyvNm6Z2Umq 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection386ＰＥＮＡＮＡficobBAHa8 維尼

Undo substitution u=√x:8964 copyright protection386ＰＥＮＡＮＡlzsFY76qkK 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection386ＰＥＮＡＮＡtfwfohHNwi 維尼

The problem is solved:8964 copyright protection386ＰＥＮＡＮＡRuQoqw4ZpX 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection386ＰＥＮＡＮＡjCaQB9PXdM 維尼

Rewrite/simplify:8964 copyright protection386ＰＥＮＡＮＡvPS1atYNNv 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection386ＰＥＮＡＮＡMu6UjdtMFH 維尼

15.158.61.37