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No Plagiarism!IKKH2CPWamrhHST2uYbdposted on PＥNAＮＡ 恐懼感8964 copyright protection541ＰＥＮＡＮＡ4UTkAMAi6u 維尼
545Please respect copyright.ＰＥＮＡＮＡQLPFG3NPHN
8964 copyright protection541ＰＥＮＡＮＡ3I07zmS8s2 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection541ＰＥＮＡＮＡiS06zOMYuI 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection541ＰＥＮＡＮＡtDDjCJuUrP 維尼
=2∫eusin(u+a)du… or choose an alternative:545Please respect copyright.ＰＥＮＡＮＡACvv4miBeT
Substitute e√x8964 copyright protection541ＰＥＮＡＮＡqj0jRk5vv9 維尼
Now solving:8964 copyright protection541ＰＥＮＡＮＡWmfGl8CZLi 維尼
∫eusin(u+a)du8964 copyright protection541ＰＥＮＡＮＡjaxueQOo3J 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection541ＰＥＮＡＮＡonPhfn41yA 維尼
First time:8964 copyright protection541ＰＥＮＡＮＡfuyJ3aGZvQ 維尼
f=sin(u+a),g′=eu8964 copyright protection541ＰＥＮＡＮＡl1FbxvRpkx 維尼
↓ steps↓ steps8964 copyright protection541ＰＥＮＡＮＡUk166YvWGx 維尼
f′=cos(u+a),g=eu:8964 copyright protection541ＰＥＮＡＮＡVvp5bEgDPt 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection541ＰＥＮＡＮＡHTPjZAtCPU 維尼
Second time:8964 copyright protection541ＰＥＮＡＮＡJdPLp1nVbw 維尼
f=cos(u+a),g′=eu8964 copyright protection541ＰＥＮＡＮＡwe2O4RNRUx 維尼
↓ steps↓ steps8964 copyright protection541ＰＥＮＡＮＡ2WgNJBu5ux 維尼
f′=−sin(u+a),g=eu:8964 copyright protection541ＰＥＮＡＮＡsBtD1QJxx4 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection541ＰＥＮＡＮＡ7a2mVhn43Z 維尼
Apply linearity:8964 copyright protection541ＰＥＮＡＮＡR5ed0NQouW 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection541ＰＥＮＡＮＡ4tULLxrBy8 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection541ＰＥＮＡＮＡ3Dw0zsdt98 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection541ＰＥＮＡＮＡUkyFN5rEDZ 維尼
Plug in solved integrals:8964 copyright protection541ＰＥＮＡＮＡ2V5yG1iD9d 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection541ＰＥＮＡＮＡDlZ0X2uIp4 維尼
Undo substitution u=√x:8964 copyright protection541ＰＥＮＡＮＡy1ThrtOmyn 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection541ＰＥＮＡＮＡ8ETGGZIuFy 維尼
The problem is solved:8964 copyright protection541ＰＥＮＡＮＡsLwcxaFtaU 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection541ＰＥＮＡＮＡPVRs6ph7ow 維尼
Rewrite/simplify:8964 copyright protection541ＰＥＮＡＮＡfqSTy7cBJb 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection541ＰＥＮＡＮＡ6t3LfQoHE4 維尼
216.73.216.173
ns216.73.216.173da2
