![]()
x
No Plagiarism!Cj0N5ezwP9CD3UbyTd0Jposted on PＥNAＮＡ 恐懼感8964 copyright protection538ＰＥＮＡＮＡtCyxGRT1nd 維尼
542Please respect copyright.ＰＥＮＡＮＡ8eMgdLYAQ1
8964 copyright protection538ＰＥＮＡＮＡbPQOHWMPge 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection538ＰＥＮＡＮＡwyZndMLllP 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection538ＰＥＮＡＮＡJO49uKIWev 維尼
=2∫eusin(u+a)du… or choose an alternative:542Please respect copyright.ＰＥＮＡＮＡBi2fKuJTSy
Substitute e√x8964 copyright protection538ＰＥＮＡＮＡiEpOEfGB7y 維尼
Now solving:8964 copyright protection538ＰＥＮＡＮＡbPDzjLKFKK 維尼
∫eusin(u+a)du8964 copyright protection538ＰＥＮＡＮＡukxiMFgeVV 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection538ＰＥＮＡＮＡmL20eHutrJ 維尼
First time:8964 copyright protection538ＰＥＮＡＮＡt1B4YUefga 維尼
f=sin(u+a),g′=eu8964 copyright protection538ＰＥＮＡＮＡGk7c4kppOZ 維尼
↓ steps↓ steps8964 copyright protection538ＰＥＮＡＮＡw3ha4JR3gO 維尼
f′=cos(u+a),g=eu:8964 copyright protection538ＰＥＮＡＮＡrEKZXjYVH5 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection538ＰＥＮＡＮＡUhzSWA7Y8U 維尼
Second time:8964 copyright protection538ＰＥＮＡＮＡLjuvGzcsKZ 維尼
f=cos(u+a),g′=eu8964 copyright protection538ＰＥＮＡＮＡ0GTPxLQw8J 維尼
↓ steps↓ steps8964 copyright protection538ＰＥＮＡＮＡTdUCeRm3Sb 維尼
f′=−sin(u+a),g=eu:8964 copyright protection538ＰＥＮＡＮＡ1jkE7D6XtJ 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection538ＰＥＮＡＮＡGdJoIRiBEn 維尼
Apply linearity:8964 copyright protection538ＰＥＮＡＮＡsVUZrkHItv 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection538ＰＥＮＡＮＡ8pJpxiY0HY 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection538ＰＥＮＡＮＡ4DtdejQhKo 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection538ＰＥＮＡＮＡTyI3475kxX 維尼
Plug in solved integrals:8964 copyright protection538ＰＥＮＡＮＡYeoZE4ZglV 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection538ＰＥＮＡＮＡ45rfZfmXt4 維尼
Undo substitution u=√x:8964 copyright protection538ＰＥＮＡＮＡjXZ5dYsvGf 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection538ＰＥＮＡＮＡvyW4zhVeWk 維尼
The problem is solved:8964 copyright protection538ＰＥＮＡＮＡNCEkKxjuYJ 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection538ＰＥＮＡＮＡYqgKKI99T6 維尼
Rewrite/simplify:8964 copyright protection538ＰＥＮＡＮＡLxIWOnRqWZ 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection538ＰＥＮＡＮＡwUA4VWtghC 維尼
216.73.216.94
ns216.73.216.94da2
