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恐懼感8964 copyright protection446ＰＥＮＡＮＡQu7q4swiVC 維尼

450Please respect copyright.ＰＥＮＡＮＡGjy6JQOOd2
8964 copyright protection446ＰＥＮＡＮＡcIiSftL5on 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection446ＰＥＮＡＮＡoTAxBVP6w6 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection446ＰＥＮＡＮＡKbb2IRUREO 維尼

=2∫eusin(u+a)du… or choose an alternative:450Please respect copyright.ＰＥＮＡＮＡ7TYNkJt9W9
Substitute e√x8964 copyright protection446ＰＥＮＡＮＡW49TqdFnhP 維尼

Now solving:8964 copyright protection446ＰＥＮＡＮＡrsyPDs0WMB 維尼

∫eusin(u+a)du8964 copyright protection446ＰＥＮＡＮＡldHsvr4492 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection446ＰＥＮＡＮＡUpzeYGZJ3D 維尼

First time:8964 copyright protection446ＰＥＮＡＮＡvtp19Ikp0A 維尼

f=sin(u+a),g′=eu8964 copyright protection446ＰＥＮＡＮＡqdmLOGcHiw 維尼

↓ steps↓ steps8964 copyright protection446ＰＥＮＡＮＡM8Cx7nW549 維尼

f′=cos(u+a),g=eu:8964 copyright protection446ＰＥＮＡＮＡvVLmIhblY1 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection446ＰＥＮＡＮＡAA8FIKwotp 維尼

Second time:8964 copyright protection446ＰＥＮＡＮＡ1lMOozPl2n 維尼

f=cos(u+a),g′=eu8964 copyright protection446ＰＥＮＡＮＡfA7s0i00YB 維尼

↓ steps↓ steps8964 copyright protection446ＰＥＮＡＮＡ44hKSx1PMl 維尼

f′=−sin(u+a),g=eu:8964 copyright protection446ＰＥＮＡＮＡnJoLuFQrbl 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection446ＰＥＮＡＮＡqDlpcSy0p4 維尼

Apply linearity:8964 copyright protection446ＰＥＮＡＮＡAzyXSdesOZ 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection446ＰＥＮＡＮＡ4eDlT5dFTX 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection446ＰＥＮＡＮＡGzFdne7FsJ 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection446ＰＥＮＡＮＡw8i1S3jNo5 維尼

Plug in solved integrals:8964 copyright protection446ＰＥＮＡＮＡVkmfvQIiE7 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection446ＰＥＮＡＮＡn2o62t029Y 維尼

Undo substitution u=√x:8964 copyright protection446ＰＥＮＡＮＡ6jXY2JiHgC 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection446ＰＥＮＡＮＡ8xdRgWqMqx 維尼

The problem is solved:8964 copyright protection446ＰＥＮＡＮＡw5GiUKVJxK 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection446ＰＥＮＡＮＡfDBLWwmsLo 維尼

Rewrite/simplify:8964 copyright protection446ＰＥＮＡＮＡ5O387znKoJ 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection446ＰＥＮＡＮＡniIsdMDOAW 維尼

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