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恐懼感8964 copyright protection403ＰＥＮＡＮＡB0WGsWSs7u 維尼

407Please respect copyright.ＰＥＮＡＮＡiEWqotfcEE
8964 copyright protection403ＰＥＮＡＮＡBUtl3NxPYx 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection403ＰＥＮＡＮＡA8vDSotwZT 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection403ＰＥＮＡＮＡS5PYTfyu9a 維尼

=2∫eusin(u+a)du… or choose an alternative:407Please respect copyright.ＰＥＮＡＮＡ1rtGXEc2Xb
Substitute e√x8964 copyright protection403ＰＥＮＡＮＡYrzwdyfl8U 維尼

Now solving:8964 copyright protection403ＰＥＮＡＮＡ1GlTpVyIqF 維尼

∫eusin(u+a)du8964 copyright protection403ＰＥＮＡＮＡsbRIcTiF75 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection403ＰＥＮＡＮＡ5GBlsbKDTD 維尼

First time:8964 copyright protection403ＰＥＮＡＮＡupBKVIo6BY 維尼

f=sin(u+a),g′=eu8964 copyright protection403ＰＥＮＡＮＡ1ba9ZKSfvc 維尼

↓ steps↓ steps8964 copyright protection403ＰＥＮＡＮＡS5nx1L8tB2 維尼

f′=cos(u+a),g=eu:8964 copyright protection403ＰＥＮＡＮＡcpMTyj3aM0 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection403ＰＥＮＡＮＡdVaktpVO9N 維尼

Second time:8964 copyright protection403ＰＥＮＡＮＡcE5yea7yXR 維尼

f=cos(u+a),g′=eu8964 copyright protection403ＰＥＮＡＮＡFCjlMxXCm4 維尼

↓ steps↓ steps8964 copyright protection403ＰＥＮＡＮＡHqHdp3K0Oh 維尼

f′=−sin(u+a),g=eu:8964 copyright protection403ＰＥＮＡＮＡ456wXVTGM7 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection403ＰＥＮＡＮＡPKDoNW6PPD 維尼

Apply linearity:8964 copyright protection403ＰＥＮＡＮＡqRMbCpkYsc 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection403ＰＥＮＡＮＡdr1af2jJeM 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection403ＰＥＮＡＮＡpb7fo4Ilzi 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection403ＰＥＮＡＮＡ6jlseMMkqT 維尼

Plug in solved integrals:8964 copyright protection403ＰＥＮＡＮＡgbyW7q0nwy 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection403ＰＥＮＡＮＡacuXQlx2LT 維尼

Undo substitution u=√x:8964 copyright protection403ＰＥＮＡＮＡgpWHw3vDcw 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection403ＰＥＮＡＮＡZNZtznsiTQ 維尼

The problem is solved:8964 copyright protection403ＰＥＮＡＮＡUq82R9Silx 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection403ＰＥＮＡＮＡIldccP0bG1 維尼

Rewrite/simplify:8964 copyright protection403ＰＥＮＡＮＡyQrxzDVa2W 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection403ＰＥＮＡＮＡW7TRU5z5sw 維尼

15.158.61.20