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恐懼感8964 copyright protection558ＰＥＮＡＮＡbXpC1eKcNh 維尼

562Please respect copyright.ＰＥＮＡＮＡKOpbPi74cY
8964 copyright protection558ＰＥＮＡＮＡCLxZD5NZDk 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection558ＰＥＮＡＮＡXNFcmH2iNq 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection558ＰＥＮＡＮＡ047VZATztF 維尼

=2∫eusin(u+a)du… or choose an alternative:562Please respect copyright.ＰＥＮＡＮＡ7h7O3ECLji
Substitute e√x8964 copyright protection558ＰＥＮＡＮＡMeNG3dVryK 維尼

Now solving:8964 copyright protection558ＰＥＮＡＮＡveyjQC4TN6 維尼

∫eusin(u+a)du8964 copyright protection558ＰＥＮＡＮＡ7qTOs6FUoy 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection558ＰＥＮＡＮＡuPokKpg4i6 維尼

First time:8964 copyright protection558ＰＥＮＡＮＡI0wcFKYDmJ 維尼

f=sin(u+a),g′=eu8964 copyright protection558ＰＥＮＡＮＡ3BKsC98zob 維尼

↓ steps↓ steps8964 copyright protection558ＰＥＮＡＮＡHiehrCZJuh 維尼

f′=cos(u+a),g=eu:8964 copyright protection558ＰＥＮＡＮＡTegAHlnTCZ 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection558ＰＥＮＡＮＡlHPNR4Km5C 維尼

Second time:8964 copyright protection558ＰＥＮＡＮＡ8dwoTyZCd1 維尼

f=cos(u+a),g′=eu8964 copyright protection558ＰＥＮＡＮＡVRLiHNF2lw 維尼

↓ steps↓ steps8964 copyright protection558ＰＥＮＡＮＡu2B5vsTMfs 維尼

f′=−sin(u+a),g=eu:8964 copyright protection558ＰＥＮＡＮＡ3MGCNhVHqO 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection558ＰＥＮＡＮＡMiuGeeT4jH 維尼

Apply linearity:8964 copyright protection558ＰＥＮＡＮＡZVIpB7C2wt 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection558ＰＥＮＡＮＡ6y1tnBUqsv 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection558ＰＥＮＡＮＡFcUKHDFiXH 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection558ＰＥＮＡＮＡSekPE1AWXa 維尼

Plug in solved integrals:8964 copyright protection558ＰＥＮＡＮＡ1Uuvz5KaxL 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection558ＰＥＮＡＮＡfXKcdLGoIj 維尼

Undo substitution u=√x:8964 copyright protection558ＰＥＮＡＮＡTpI85q0koh 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection558ＰＥＮＡＮＡRNNxpnIOY2 維尼

The problem is solved:8964 copyright protection558ＰＥＮＡＮＡZHI0tbAX0U 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection558ＰＥＮＡＮＡByY1s1jQO9 維尼

Rewrite/simplify:8964 copyright protection558ＰＥＮＡＮＡZHiemrgzDC 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection558ＰＥＮＡＮＡRA4wK0Fe2P 維尼

216.73.216.12