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No Plagiarism!fJD3lbzVUDxU9B6jNO3jposted on PＥNAＮＡ 恐懼感8964 copyright protection466ＰＥＮＡＮＡyYQw4DCRgi 維尼
470Please respect copyright.ＰＥＮＡＮＡnwsesdIjjp
8964 copyright protection466ＰＥＮＡＮＡDmERqRR6pH 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection466ＰＥＮＡＮＡW04EuNnoeQ 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection466ＰＥＮＡＮＡ1Eqllf7AHV 維尼
=2∫eusin(u+a)du… or choose an alternative:470Please respect copyright.ＰＥＮＡＮＡQCqWzfneNT
Substitute e√x8964 copyright protection466ＰＥＮＡＮＡIFJoby5fyY 維尼
Now solving:8964 copyright protection466ＰＥＮＡＮＡUlJOOm7Ugp 維尼
∫eusin(u+a)du8964 copyright protection466ＰＥＮＡＮＡKLNTQx95yE 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection466ＰＥＮＡＮＡDQuQjAOSU6 維尼
First time:8964 copyright protection466ＰＥＮＡＮＡal8mu4oxsB 維尼
f=sin(u+a),g′=eu8964 copyright protection466ＰＥＮＡＮＡA5NIqzLxi0 維尼
↓ steps↓ steps8964 copyright protection466ＰＥＮＡＮＡJMtsMRuzfz 維尼
f′=cos(u+a),g=eu:8964 copyright protection466ＰＥＮＡＮＡuQYfhi5fCu 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection466ＰＥＮＡＮＡAuTit04slG 維尼
Second time:8964 copyright protection466ＰＥＮＡＮＡe5DEWPycoA 維尼
f=cos(u+a),g′=eu8964 copyright protection466ＰＥＮＡＮＡIKWsiVcwA4 維尼
↓ steps↓ steps8964 copyright protection466ＰＥＮＡＮＡXMEibE1Yag 維尼
f′=−sin(u+a),g=eu:8964 copyright protection466ＰＥＮＡＮＡ2COhHIoRa1 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection466ＰＥＮＡＮＡFNdWRceGjD 維尼
Apply linearity:8964 copyright protection466ＰＥＮＡＮＡ6vtQD1mbor 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection466ＰＥＮＡＮＡmsTIY1iKFa 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection466ＰＥＮＡＮＡIfsN0wPB9I 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection466ＰＥＮＡＮＡKFMj5CWd2A 維尼
Plug in solved integrals:8964 copyright protection466ＰＥＮＡＮＡnwcyJ9MzzO 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection466ＰＥＮＡＮＡa38m7JiQqj 維尼
Undo substitution u=√x:8964 copyright protection466ＰＥＮＡＮＡXVTKxWuPKR 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection466ＰＥＮＡＮＡ4f0ojCvBFX 維尼
The problem is solved:8964 copyright protection466ＰＥＮＡＮＡ0dEeNOOot9 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection466ＰＥＮＡＮＡ0g9bxETlLW 維尼
Rewrite/simplify:8964 copyright protection466ＰＥＮＡＮＡdQMZNRpEp3 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection466ＰＥＮＡＮＡP2xpdDeBi9 維尼
13.59.134.12
ns13.59.134.12da2
