No Plagiarism!BIoLhpRlwWnokDfkHL9fposted on PENANA 恐懼感8964 copyright protection465PENANAusdlhgRXF2 維尼
469Please respect copyright.PENANAFIb4OuwjQ8
8964 copyright protection465PENANARHRXBxDhqD 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection465PENANAnV9a5ViDZU 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection465PENANAeSwe4fpmCx 維尼
=2∫eusin(u+a)du… or choose an alternative:469Please respect copyright.PENANA7fNXkJj4Tn
Substitute e√x8964 copyright protection465PENANATtMzrXkka0 維尼
Now solving:8964 copyright protection465PENANAScKvQm7cQy 維尼
∫eusin(u+a)du8964 copyright protection465PENANA6G2WeKkbtj 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection465PENANAtWPOmzW3GW 維尼
First time:8964 copyright protection465PENANAVHNHP9DgfT 維尼
f=sin(u+a),g′=eu8964 copyright protection465PENANAUeg2w7rhKe 維尼
↓ steps↓ steps8964 copyright protection465PENANA1e2Px2EJPe 維尼
f′=cos(u+a),g=eu:8964 copyright protection465PENANAifXVpJdTV8 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection465PENANAPoz72Krpvs 維尼
Second time:8964 copyright protection465PENANAgCSof4kGvZ 維尼
f=cos(u+a),g′=eu8964 copyright protection465PENANAr2gZnBHnO4 維尼
↓ steps↓ steps8964 copyright protection465PENANA9avjyQI4b2 維尼
f′=−sin(u+a),g=eu:8964 copyright protection465PENANAWchuvHf5Ur 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection465PENANAbxcsmPcLFM 維尼
Apply linearity:8964 copyright protection465PENANAeRlKx58dEY 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection465PENANAbxtUEDcgPb 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection465PENANAD2J2i9Qv7N 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection465PENANAGMiwQHTvi3 維尼
Plug in solved integrals:8964 copyright protection465PENANAC8yvdhatAI 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection465PENANAMu7S6o23et 維尼
Undo substitution u=√x:8964 copyright protection465PENANAypI4pYFccb 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection465PENANAh5TgKix0iD 維尼
The problem is solved:8964 copyright protection465PENANABY7S2yLjyX 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection465PENANAM8xuWKusWq 維尼
Rewrite/simplify:8964 copyright protection465PENANANWKDBKT97l 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection465PENANA70K7o1Tp17 維尼
18.227.49.178
ns18.227.49.178da2