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恐懼感8964 copyright protection478ＰＥＮＡＮＡsoZP1gaTdZ 維尼

482Please respect copyright.ＰＥＮＡＮＡ0u9rmZ3TGw
8964 copyright protection478ＰＥＮＡＮＡyHLq13U7sF 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection478ＰＥＮＡＮＡFHzN8B1MNE 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection478ＰＥＮＡＮＡTMcsLTOxWR 維尼

=2∫eusin(u+a)du… or choose an alternative:482Please respect copyright.ＰＥＮＡＮＡYYeEOb71Ri
Substitute e√x8964 copyright protection478ＰＥＮＡＮＡutbFtg702P 維尼

Now solving:8964 copyright protection478ＰＥＮＡＮＡgjyzMpDFzk 維尼

∫eusin(u+a)du8964 copyright protection478ＰＥＮＡＮＡ9G0qQrzBLN 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection478ＰＥＮＡＮＡSLlCgceXM7 維尼

First time:8964 copyright protection478ＰＥＮＡＮＡeuqEP7BimA 維尼

f=sin(u+a),g′=eu8964 copyright protection478ＰＥＮＡＮＡ1H0md4Gq5K 維尼

↓ steps↓ steps8964 copyright protection478ＰＥＮＡＮＡMtqiYbPGcb 維尼

f′=cos(u+a),g=eu:8964 copyright protection478ＰＥＮＡＮＡPXS2nm9HTf 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection478ＰＥＮＡＮＡfL3GJVpGvB 維尼

Second time:8964 copyright protection478ＰＥＮＡＮＡF3VnqRvfbv 維尼

f=cos(u+a),g′=eu8964 copyright protection478ＰＥＮＡＮＡxvoL2gfdes 維尼

↓ steps↓ steps8964 copyright protection478ＰＥＮＡＮＡTRKiaNdEKy 維尼

f′=−sin(u+a),g=eu:8964 copyright protection478ＰＥＮＡＮＡxP6upg3z2G 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection478ＰＥＮＡＮＡ3UgxZsh37c 維尼

Apply linearity:8964 copyright protection478ＰＥＮＡＮＡXtyWWIe5q9 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection478ＰＥＮＡＮＡwn13ge23qk 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection478ＰＥＮＡＮＡvykK9hyY9p 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection478ＰＥＮＡＮＡz45NeUBLVm 維尼

Plug in solved integrals:8964 copyright protection478ＰＥＮＡＮＡ2sYEvTbEYD 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection478ＰＥＮＡＮＡoynTS20btC 維尼

Undo substitution u=√x:8964 copyright protection478ＰＥＮＡＮＡ4j846qQ4hk 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection478ＰＥＮＡＮＡ3L0qOJ36bl 維尼

The problem is solved:8964 copyright protection478ＰＥＮＡＮＡaBHxiAPiuN 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection478ＰＥＮＡＮＡ1DKt1RRH21 維尼

Rewrite/simplify:8964 copyright protection478ＰＥＮＡＮＡhceftPWaXL 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection478ＰＥＮＡＮＡ7YAMK5ETv8 維尼

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