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恐懼感8964 copyright protection437ＰＥＮＡＮＡETWwCNcy4h 維尼

441Please respect copyright.ＰＥＮＡＮＡR5UAuQVd3h
8964 copyright protection437ＰＥＮＡＮＡ8MWJ2mcvQa 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection437ＰＥＮＡＮＡsxf2aXa9Af 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection437ＰＥＮＡＮＡFyOkJngyxT 維尼

=2∫eusin(u+a)du… or choose an alternative:441Please respect copyright.ＰＥＮＡＮＡQYqGR68Kmy
Substitute e√x8964 copyright protection437ＰＥＮＡＮＡEHkhUSxTIX 維尼

Now solving:8964 copyright protection437ＰＥＮＡＮＡ9EIsB6tQcq 維尼

∫eusin(u+a)du8964 copyright protection437ＰＥＮＡＮＡKQ6QMzx8BK 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection437ＰＥＮＡＮＡPOTChm3gZx 維尼

First time:8964 copyright protection437ＰＥＮＡＮＡqwubha1tFx 維尼

f=sin(u+a),g′=eu8964 copyright protection437ＰＥＮＡＮＡ2Yg5Myuf4s 維尼

↓ steps↓ steps8964 copyright protection437ＰＥＮＡＮＡwlyiq0l44R 維尼

f′=cos(u+a),g=eu:8964 copyright protection437ＰＥＮＡＮＡvkOZ73pdJS 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection437ＰＥＮＡＮＡZRSoLwFSfX 維尼

Second time:8964 copyright protection437ＰＥＮＡＮＡx4xCDl22dF 維尼

f=cos(u+a),g′=eu8964 copyright protection437ＰＥＮＡＮＡgWUuTl7CCy 維尼

↓ steps↓ steps8964 copyright protection437ＰＥＮＡＮＡNog45WRoEH 維尼

f′=−sin(u+a),g=eu:8964 copyright protection437ＰＥＮＡＮＡg27MEIPI7Q 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection437ＰＥＮＡＮＡc5BuoCKwrr 維尼

Apply linearity:8964 copyright protection437ＰＥＮＡＮＡINdcUYNmmK 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection437ＰＥＮＡＮＡcT2KxLhAdw 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection437ＰＥＮＡＮＡxo4o31geyP 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection437ＰＥＮＡＮＡDuJTLAAKP8 維尼

Plug in solved integrals:8964 copyright protection437ＰＥＮＡＮＡyG3DFCj5nk 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection437ＰＥＮＡＮＡ6ZcNkxzyy1 維尼

Undo substitution u=√x:8964 copyright protection437ＰＥＮＡＮＡkKn3B1O9Wf 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection437ＰＥＮＡＮＡiFL1g0ft8j 維尼

The problem is solved:8964 copyright protection437ＰＥＮＡＮＡ28BezXLmfS 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection437ＰＥＮＡＮＡlP6SHHjxOJ 維尼

Rewrite/simplify:8964 copyright protection437ＰＥＮＡＮＡJfs5MA7ZOL 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection437ＰＥＮＡＮＡTKWahg3eOA 維尼

3.131.38.100