No Plagiarism!ocLhBD1M1YdnyPXci30Dposted on PENANA 恐懼感8964 copyright protection295PENANAvA38ukqKfc 維尼
299Please respect copyright.PENANAGA0E2gsZak
8964 copyright protection295PENANA13vPdVnjNL 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection295PENANAjK6yn7POKw 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection295PENANAoQxjMlgAxw 維尼
=2∫eusin(u+a)du… or choose an alternative:299Please respect copyright.PENANASdajvjdFRF
Substitute e√x8964 copyright protection295PENANAXx296pJL3k 維尼
Now solving:8964 copyright protection295PENANAxAKx5f4jjJ 維尼
∫eusin(u+a)du8964 copyright protection295PENANA7qnUpE8Cnu 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection295PENANA2WrWoiHKYE 維尼
First time:8964 copyright protection295PENANAJmc9htKEDR 維尼
f=sin(u+a),g′=eu8964 copyright protection295PENANAEMzfdd5sYf 維尼
↓ steps↓ steps8964 copyright protection295PENANA6njTgBpgpI 維尼
f′=cos(u+a),g=eu:8964 copyright protection295PENANAGCiHiIoWGx 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection295PENANAzUD2vLJxqD 維尼
Second time:8964 copyright protection295PENANA5NtUmXbPo1 維尼
f=cos(u+a),g′=eu8964 copyright protection295PENANAJtAMBjtK0F 維尼
↓ steps↓ steps8964 copyright protection295PENANAk2kQwGHJIw 維尼
f′=−sin(u+a),g=eu:8964 copyright protection295PENANAZtqJvUtgPv 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection295PENANAtWzFueNFTv 維尼
Apply linearity:8964 copyright protection295PENANAKiOYjPCUgA 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection295PENANA6H6KBoZgUT 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection295PENANAkOTBLMAtKd 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection295PENANA53wz8WgdrD 維尼
Plug in solved integrals:8964 copyright protection295PENANAQUcUy3RRYG 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection295PENANAbsr6HfzI11 維尼
Undo substitution u=√x:8964 copyright protection295PENANAApLdQYEueV 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection295PENANApFVFvHrV4F 維尼
The problem is solved:8964 copyright protection295PENANAyUJzg3sD2z 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection295PENANANAFsdS88wL 維尼
Rewrite/simplify:8964 copyright protection295PENANApWye3oPadv 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection295PENANAi42PvVKe9Y 維尼
172.70.126.20
ns 172.70.126.20da2