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No Plagiarism!NLVNatUr3uTD1Nke12c2posted on PＥNAＮＡ 恐懼感8964 copyright protection468ＰＥＮＡＮＡ3llldDrTNf 維尼
472Please respect copyright.ＰＥＮＡＮＡNkf3Pm0hRe
8964 copyright protection468ＰＥＮＡＮＡxze0oZNWqg 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection468ＰＥＮＡＮＡIsjVQbFX0K 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection468ＰＥＮＡＮＡfvaxdlLn7W 維尼
=2∫eusin(u+a)du… or choose an alternative:472Please respect copyright.ＰＥＮＡＮＡOCFs8HCzX2
Substitute e√x8964 copyright protection468ＰＥＮＡＮＡsLZd5aDrE5 維尼
Now solving:8964 copyright protection468ＰＥＮＡＮＡ56RF3x2meK 維尼
∫eusin(u+a)du8964 copyright protection468ＰＥＮＡＮＡhJcIhA5FbG 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection468ＰＥＮＡＮＡCJoWzGu1rp 維尼
First time:8964 copyright protection468ＰＥＮＡＮＡmH058dhYnE 維尼
f=sin(u+a),g′=eu8964 copyright protection468ＰＥＮＡＮＡvCvQQOEEpH 維尼
↓ steps↓ steps8964 copyright protection468ＰＥＮＡＮＡGxqnOIQ6OT 維尼
f′=cos(u+a),g=eu:8964 copyright protection468ＰＥＮＡＮＡy4PYu02Ucd 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection468ＰＥＮＡＮＡ6xUB3yKfP9 維尼
Second time:8964 copyright protection468ＰＥＮＡＮＡblnTmSHDlS 維尼
f=cos(u+a),g′=eu8964 copyright protection468ＰＥＮＡＮＡI8NX5NaZQ1 維尼
↓ steps↓ steps8964 copyright protection468ＰＥＮＡＮＡDBAh6QHzUe 維尼
f′=−sin(u+a),g=eu:8964 copyright protection468ＰＥＮＡＮＡGa2NrBjQDK 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection468ＰＥＮＡＮＡg23gZcIRvq 維尼
Apply linearity:8964 copyright protection468ＰＥＮＡＮＡhiwtnnA8DU 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection468ＰＥＮＡＮＡrlKq7VEgMg 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection468ＰＥＮＡＮＡLdNCyicOe5 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection468ＰＥＮＡＮＡeNb9GZHXhR 維尼
Plug in solved integrals:8964 copyright protection468ＰＥＮＡＮＡ0KfiDLdk4u 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection468ＰＥＮＡＮＡyM6HQCtoeu 維尼
Undo substitution u=√x:8964 copyright protection468ＰＥＮＡＮＡMMAU1AO8LZ 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection468ＰＥＮＡＮＡ1Lp7YbYHJ5 維尼
The problem is solved:8964 copyright protection468ＰＥＮＡＮＡZLRYD285Gj 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection468ＰＥＮＡＮＡp2NFo3TZCC 維尼
Rewrite/simplify:8964 copyright protection468ＰＥＮＡＮＡqc6t2i7WNZ 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection468ＰＥＮＡＮＡ28AcZ1v1KD 維尼
18.119.121.38
ns18.119.121.38da2
