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No Plagiarism!BJZ0AC8Or0Y9o5OWnxdqposted on PＥNAＮＡ 恐懼感8964 copyright protection454ＰＥＮＡＮＡQyX7PA0vqw 維尼
458Please respect copyright.ＰＥＮＡＮＡKjI3Wz8m9y
8964 copyright protection454ＰＥＮＡＮＡcEdwcZBxOW 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection454ＰＥＮＡＮＡo4Y4GauMgy 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection454ＰＥＮＡＮＡKlqmsInyGz 維尼
=2∫eusin(u+a)du… or choose an alternative:458Please respect copyright.ＰＥＮＡＮＡW5RRG4VHh4
Substitute e√x8964 copyright protection454ＰＥＮＡＮＡZyWNRzHhNO 維尼
Now solving:8964 copyright protection454ＰＥＮＡＮＡSp06cSH8vi 維尼
∫eusin(u+a)du8964 copyright protection454ＰＥＮＡＮＡDaj1sXpOzN 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection454ＰＥＮＡＮＡV8Tl3A5ewK 維尼
First time:8964 copyright protection454ＰＥＮＡＮＡG2dXjvSFv4 維尼
f=sin(u+a),g′=eu8964 copyright protection454ＰＥＮＡＮＡCNgMeV5QaX 維尼
↓ steps↓ steps8964 copyright protection454ＰＥＮＡＮＡNdFLacG1k2 維尼
f′=cos(u+a),g=eu:8964 copyright protection454ＰＥＮＡＮＡ3e51ZC7AB2 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection454ＰＥＮＡＮＡ8kH0u66TgO 維尼
Second time:8964 copyright protection454ＰＥＮＡＮＡVzwL91ziEB 維尼
f=cos(u+a),g′=eu8964 copyright protection454ＰＥＮＡＮＡ7z3nT3uKaw 維尼
↓ steps↓ steps8964 copyright protection454ＰＥＮＡＮＡZC5fV4jYWO 維尼
f′=−sin(u+a),g=eu:8964 copyright protection454ＰＥＮＡＮＡkN1FpnimTz 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection454ＰＥＮＡＮＡb2JOu0jW3e 維尼
Apply linearity:8964 copyright protection454ＰＥＮＡＮＡH00VDYLTs2 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection454ＰＥＮＡＮＡ0AFmKvJ0mT 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection454ＰＥＮＡＮＡone54aUB8t 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection454ＰＥＮＡＮＡFsyy6YECFV 維尼
Plug in solved integrals:8964 copyright protection454ＰＥＮＡＮＡj5XDdy31T4 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection454ＰＥＮＡＮＡnpnPmh9Jqt 維尼
Undo substitution u=√x:8964 copyright protection454ＰＥＮＡＮＡI9Km8xlYJp 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection454ＰＥＮＡＮＡqTFHz1Mx7L 維尼
The problem is solved:8964 copyright protection454ＰＥＮＡＮＡ6xZoYviDkx 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection454ＰＥＮＡＮＡyS9iQvDEKz 維尼
Rewrite/simplify:8964 copyright protection454ＰＥＮＡＮＡr7C1bTo7gT 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection454ＰＥＮＡＮＡT6XE585Irk 維尼
3.140.186.140
ns3.140.186.140da2
