No Plagiarism!EzjXQxeaHzXdELqGslGxposted on PENANA 恐懼感8964 copyright protection518PENANAcvhjeZ1U3W 維尼
522Please respect copyright.PENANALQcnjB6q5Y
8964 copyright protection518PENANAIYGrUqKYaY 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection518PENANA4zB6isN79w 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection518PENANAXo0E595Vwn 維尼
=2∫eusin(u+a)du… or choose an alternative:522Please respect copyright.PENANAJvtGZSBdMM
Substitute e√x8964 copyright protection518PENANAfku3qXpSOx 維尼
Now solving:8964 copyright protection518PENANAIz60cSe6ZT 維尼
∫eusin(u+a)du8964 copyright protection518PENANAQ68YniajDw 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection518PENANAm2tvRK46HK 維尼
First time:8964 copyright protection518PENANAd6JFDqAlGR 維尼
f=sin(u+a),g′=eu8964 copyright protection518PENANArGbaBYue2u 維尼
↓ steps↓ steps8964 copyright protection518PENANAGBas8HS6v3 維尼
f′=cos(u+a),g=eu:8964 copyright protection518PENANAdVj4PwIZdv 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection518PENANAkWBfR47mae 維尼
Second time:8964 copyright protection518PENANALGP33KvPCG 維尼
f=cos(u+a),g′=eu8964 copyright protection518PENANAO3TusSL8E5 維尼
↓ steps↓ steps8964 copyright protection518PENANASiBpOtD1TT 維尼
f′=−sin(u+a),g=eu:8964 copyright protection518PENANA9MrD7IVcAD 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection518PENANAxrXcJkHzpF 維尼
Apply linearity:8964 copyright protection518PENANAGS2Jg8guxg 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection518PENANAnaR0Qe3MWE 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection518PENANAUNyIBgmO0v 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection518PENANAlQN0k40o6I 維尼
Plug in solved integrals:8964 copyright protection518PENANAy3Da6EFGA9 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection518PENANARcqALzehM4 維尼
Undo substitution u=√x:8964 copyright protection518PENANAOX61ZtTijr 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection518PENANAWuxORkgIVG 維尼
The problem is solved:8964 copyright protection518PENANAYIgS7YHLTT 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection518PENANAayyjiQLq94 維尼
Rewrite/simplify:8964 copyright protection518PENANA1Akb8oaQbK 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection518PENANAGb7vYGCBvp 維尼
216.73.216.30
ns216.73.216.30da2