![]()
x
No Plagiarism!Qh3sxOXvoyKfnyKepvhNposted on PＥNAＮＡ 恐懼感8964 copyright protection530ＰＥＮＡＮＡGCCq1uuLyD 維尼
534Please respect copyright.ＰＥＮＡＮＡXmqTmapCOS
8964 copyright protection530ＰＥＮＡＮＡ59G1bDUrlL 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection530ＰＥＮＡＮＡe1Fgv6fz0T 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection530ＰＥＮＡＮＡkZpgWlrUaZ 維尼
=2∫eusin(u+a)du… or choose an alternative:534Please respect copyright.ＰＥＮＡＮＡYpT8rUxpRd
Substitute e√x8964 copyright protection530ＰＥＮＡＮＡv0lc2zWpK9 維尼
Now solving:8964 copyright protection530ＰＥＮＡＮＡkb2Z7pWoiV 維尼
∫eusin(u+a)du8964 copyright protection530ＰＥＮＡＮＡSCPRbGwV0Q 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection530ＰＥＮＡＮＡCOChmtPNIR 維尼
First time:8964 copyright protection530ＰＥＮＡＮＡRVGbemhJVG 維尼
f=sin(u+a),g′=eu8964 copyright protection530ＰＥＮＡＮＡh94n9vVDZi 維尼
↓ steps↓ steps8964 copyright protection530ＰＥＮＡＮＡAVBftsYp5h 維尼
f′=cos(u+a),g=eu:8964 copyright protection530ＰＥＮＡＮＡKQmUICATVq 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection530ＰＥＮＡＮＡK5Q2Xcw6Oe 維尼
Second time:8964 copyright protection530ＰＥＮＡＮＡEf8NsXZEf2 維尼
f=cos(u+a),g′=eu8964 copyright protection530ＰＥＮＡＮＡMllBHkwYGJ 維尼
↓ steps↓ steps8964 copyright protection530ＰＥＮＡＮＡbFn9nFmB4S 維尼
f′=−sin(u+a),g=eu:8964 copyright protection530ＰＥＮＡＮＡLKJaYdDPia 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection530ＰＥＮＡＮＡoD3DpqiZjH 維尼
Apply linearity:8964 copyright protection530ＰＥＮＡＮＡQYL6f1FOSZ 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection530ＰＥＮＡＮＡGsCrMiWC4W 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection530ＰＥＮＡＮＡHQBGgXnyHM 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection530ＰＥＮＡＮＡdvOKQYe7WR 維尼
Plug in solved integrals:8964 copyright protection530ＰＥＮＡＮＡdoBhWfCIER 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection530ＰＥＮＡＮＡ8QBnN4L792 維尼
Undo substitution u=√x:8964 copyright protection530ＰＥＮＡＮＡTodplNpKy2 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection530ＰＥＮＡＮＡuzzE87Hbzx 維尼
The problem is solved:8964 copyright protection530ＰＥＮＡＮＡ4MSLjNyC8F 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection530ＰＥＮＡＮＡr0XHVD8KSL 維尼
Rewrite/simplify:8964 copyright protection530ＰＥＮＡＮＡ6hYZn4oosA 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection530ＰＥＮＡＮＡG8A0nGD49S 維尼
216.73.216.176
ns216.73.216.176da2
