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恐懼感8964 copyright protection439ＰＥＮＡＮＡVTZwVjnqNb 維尼

443Please respect copyright.ＰＥＮＡＮＡEKiTyIHxDr
8964 copyright protection439ＰＥＮＡＮＡuTVXaODXaK 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection439ＰＥＮＡＮＡKL0YA57zUO 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection439ＰＥＮＡＮＡDaevSVZToA 維尼

=2∫eusin(u+a)du… or choose an alternative:443Please respect copyright.ＰＥＮＡＮＡzyGIk4BTh0
Substitute e√x8964 copyright protection439ＰＥＮＡＮＡFOA7uvSAuh 維尼

Now solving:8964 copyright protection439ＰＥＮＡＮＡToZJdPbB0J 維尼

∫eusin(u+a)du8964 copyright protection439ＰＥＮＡＮＡda1A5wBCi7 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection439ＰＥＮＡＮＡ609RSGyvXh 維尼

First time:8964 copyright protection439ＰＥＮＡＮＡDquab4LWgy 維尼

f=sin(u+a),g′=eu8964 copyright protection439ＰＥＮＡＮＡRf7K59J8nj 維尼

↓ steps↓ steps8964 copyright protection439ＰＥＮＡＮＡicKHL4N8KJ 維尼

f′=cos(u+a),g=eu:8964 copyright protection439ＰＥＮＡＮＡirTOS6pVtd 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection439ＰＥＮＡＮＡWnSm3VC90d 維尼

Second time:8964 copyright protection439ＰＥＮＡＮＡ4EQfT1AvHI 維尼

f=cos(u+a),g′=eu8964 copyright protection439ＰＥＮＡＮＡUX6ZNi4Kcs 維尼

↓ steps↓ steps8964 copyright protection439ＰＥＮＡＮＡo728wRxoqs 維尼

f′=−sin(u+a),g=eu:8964 copyright protection439ＰＥＮＡＮＡEm5u4f6Ajn 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection439ＰＥＮＡＮＡ3emm8vXqQA 維尼

Apply linearity:8964 copyright protection439ＰＥＮＡＮＡeWoNyeg1Wc 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection439ＰＥＮＡＮＡ4hPrcCh1i7 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection439ＰＥＮＡＮＡzDX5prbHRK 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection439ＰＥＮＡＮＡBex5rME9lf 維尼

Plug in solved integrals:8964 copyright protection439ＰＥＮＡＮＡJGfOGBLURa 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection439ＰＥＮＡＮＡvbRBPMAWOr 維尼

Undo substitution u=√x:8964 copyright protection439ＰＥＮＡＮＡTlp9a6qAxk 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection439ＰＥＮＡＮＡJItNfF9zAI 維尼

The problem is solved:8964 copyright protection439ＰＥＮＡＮＡsws9PxcFEx 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection439ＰＥＮＡＮＡPAXvUFw2RD 維尼

Rewrite/simplify:8964 copyright protection439ＰＥＮＡＮＡXZUSRvFaaG 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection439ＰＥＮＡＮＡDMPylON3uU 維尼

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