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No Plagiarism!eb8OQR8QiDi1WkPfdaiFposted on PＥNAＮＡ 恐懼感8964 copyright protection470ＰＥＮＡＮＡdY64t7dGb0 維尼
474Please respect copyright.ＰＥＮＡＮＡSHamL6oJeY
8964 copyright protection470ＰＥＮＡＮＡ5yCEnRLwF7 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection470ＰＥＮＡＮＡHIy7LhZCKM 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection470ＰＥＮＡＮＡXkjvVZyQai 維尼
=2∫eusin(u+a)du… or choose an alternative:474Please respect copyright.ＰＥＮＡＮＡ4lzMadiEdd
Substitute e√x8964 copyright protection470ＰＥＮＡＮＡj3f7HrGPu6 維尼
Now solving:8964 copyright protection470ＰＥＮＡＮＡKS4OD7Zqj9 維尼
∫eusin(u+a)du8964 copyright protection470ＰＥＮＡＮＡK2X09jEsVs 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection470ＰＥＮＡＮＡdA1rddwYJO 維尼
First time:8964 copyright protection470ＰＥＮＡＮＡWqLUuBQKRO 維尼
f=sin(u+a),g′=eu8964 copyright protection470ＰＥＮＡＮＡKzmWNERkjE 維尼
↓ steps↓ steps8964 copyright protection470ＰＥＮＡＮＡiA7AKDlHxE 維尼
f′=cos(u+a),g=eu:8964 copyright protection470ＰＥＮＡＮＡGTfrYbHdpX 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection470ＰＥＮＡＮＡpi5hwlvlco 維尼
Second time:8964 copyright protection470ＰＥＮＡＮＡc50rckUMgK 維尼
f=cos(u+a),g′=eu8964 copyright protection470ＰＥＮＡＮＡ4nuk3L8LiF 維尼
↓ steps↓ steps8964 copyright protection470ＰＥＮＡＮＡIAWqNJsXOt 維尼
f′=−sin(u+a),g=eu:8964 copyright protection470ＰＥＮＡＮＡAZMLcTbycn 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection470ＰＥＮＡＮＡE2JuxjmTkC 維尼
Apply linearity:8964 copyright protection470ＰＥＮＡＮＡ88XCsIw6FL 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection470ＰＥＮＡＮＡrI40xyi4pQ 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection470ＰＥＮＡＮＡHTGElDsHTE 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection470ＰＥＮＡＮＡqZrM3YdKXB 維尼
Plug in solved integrals:8964 copyright protection470ＰＥＮＡＮＡ3K0xetCp5s 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection470ＰＥＮＡＮＡ2WzcCaJ34i 維尼
Undo substitution u=√x:8964 copyright protection470ＰＥＮＡＮＡv9ZCaqrNTU 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection470ＰＥＮＡＮＡqRn0cp2wXK 維尼
The problem is solved:8964 copyright protection470ＰＥＮＡＮＡATUDE6YOCF 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection470ＰＥＮＡＮＡbe28Ex4EH4 維尼
Rewrite/simplify:8964 copyright protection470ＰＥＮＡＮＡ5QEDe9tWhX 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection470ＰＥＮＡＮＡ1O0JueZZ93 維尼
18.118.104.28
ns18.118.104.28da2
