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No Plagiarism!tO31YeetpM1YX7fperpKposted on PＥNAＮＡ 恐懼感8964 copyright protection525ＰＥＮＡＮＡH9d23sa7px 維尼
529Please respect copyright.ＰＥＮＡＮＡ33EVrI911m
8964 copyright protection525ＰＥＮＡＮＡllxYRXOlMQ 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection525ＰＥＮＡＮＡblw6QKS1Iq 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection525ＰＥＮＡＮＡVWiPAxUDnK 維尼
=2∫eusin(u+a)du… or choose an alternative:529Please respect copyright.ＰＥＮＡＮＡKTX4VgdcZj
Substitute e√x8964 copyright protection525ＰＥＮＡＮＡobqpViippk 維尼
Now solving:8964 copyright protection525ＰＥＮＡＮＡ66ZHDMJOdV 維尼
∫eusin(u+a)du8964 copyright protection525ＰＥＮＡＮＡugsg2XZ6ZI 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection525ＰＥＮＡＮＡczEKu0SS54 維尼
First time:8964 copyright protection525ＰＥＮＡＮＡ5FjNxZvjo5 維尼
f=sin(u+a),g′=eu8964 copyright protection525ＰＥＮＡＮＡv2CIT3K2td 維尼
↓ steps↓ steps8964 copyright protection525ＰＥＮＡＮＡX7XscSaD4R 維尼
f′=cos(u+a),g=eu:8964 copyright protection525ＰＥＮＡＮＡC5huTuiR1I 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection525ＰＥＮＡＮＡfAkhUpqd5m 維尼
Second time:8964 copyright protection525ＰＥＮＡＮＡXSlRrHRKRH 維尼
f=cos(u+a),g′=eu8964 copyright protection525ＰＥＮＡＮＡVppQjfEo9h 維尼
↓ steps↓ steps8964 copyright protection525ＰＥＮＡＮＡZYBWtZG7PI 維尼
f′=−sin(u+a),g=eu:8964 copyright protection525ＰＥＮＡＮＡI5Fbrb72te 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection525ＰＥＮＡＮＡeet5cD6WoT 維尼
Apply linearity:8964 copyright protection525ＰＥＮＡＮＡ1WpoRpZLaE 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection525ＰＥＮＡＮＡdBVUV15POe 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection525ＰＥＮＡＮＡPCUehXSKcG 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection525ＰＥＮＡＮＡ3GGGPZVtjQ 維尼
Plug in solved integrals:8964 copyright protection525ＰＥＮＡＮＡdv8ZN7qlR9 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection525ＰＥＮＡＮＡi0Gec92wAp 維尼
Undo substitution u=√x:8964 copyright protection525ＰＥＮＡＮＡymAwIdMNWX 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection525ＰＥＮＡＮＡfhlFY0iwt9 維尼
The problem is solved:8964 copyright protection525ＰＥＮＡＮＡDJgM5yGeRK 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection525ＰＥＮＡＮＡgYJbz3ZKwP 維尼
Rewrite/simplify:8964 copyright protection525ＰＥＮＡＮＡmkErDhf3ky 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection525ＰＥＮＡＮＡKaZsSVRirT 維尼
216.73.216.146
ns216.73.216.146da2
