No Plagiarism!DPgnpTDVa9yUKiJzG6sTposted on PENANA 恐懼感8964 copyright protection467PENANA3oQVBbcRTi 維尼
471Please respect copyright.PENANAsEnD8e5zbv
8964 copyright protection467PENANAq4Y8HC3X99 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection467PENANAODV58CEsfs 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection467PENANA21iJM2HQ2Y 維尼
=2∫eusin(u+a)du… or choose an alternative:471Please respect copyright.PENANAB0Xy3JUStr
Substitute e√x8964 copyright protection467PENANA6G1mqXSYFi 維尼
Now solving:8964 copyright protection467PENANAtE40H8a4cR 維尼
∫eusin(u+a)du8964 copyright protection467PENANAkEn1IgYwmq 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection467PENANAWuyByO0aAp 維尼
First time:8964 copyright protection467PENANArzJvNrozhI 維尼
f=sin(u+a),g′=eu8964 copyright protection467PENANALVg6O7Txgd 維尼
↓ steps↓ steps8964 copyright protection467PENANAJj7o7whZ9R 維尼
f′=cos(u+a),g=eu:8964 copyright protection467PENANAPHzDI312ZD 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection467PENANAg6htquIquR 維尼
Second time:8964 copyright protection467PENANAByj6SqMiuK 維尼
f=cos(u+a),g′=eu8964 copyright protection467PENANA86CicNWIVf 維尼
↓ steps↓ steps8964 copyright protection467PENANAMpOACoXmth 維尼
f′=−sin(u+a),g=eu:8964 copyright protection467PENANAG07rGGlV2E 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection467PENANAxMW4ktnmw4 維尼
Apply linearity:8964 copyright protection467PENANAT3lv88a2WX 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection467PENANA94iGTSWQEF 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection467PENANASlTMIxVolv 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection467PENANAjFiYRJofvH 維尼
Plug in solved integrals:8964 copyright protection467PENANA0jY9ss6Sph 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection467PENANA96hi2kikA7 維尼
Undo substitution u=√x:8964 copyright protection467PENANA8Rr8aH5ZKT 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection467PENANAu65l3qA3sQ 維尼
The problem is solved:8964 copyright protection467PENANA5k5SDnThvN 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection467PENANAOfQyglyUoS 維尼
Rewrite/simplify:8964 copyright protection467PENANAgdwxeZ8GdU 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection467PENANAO8KvcS2BiO 維尼
18.117.158.108
ns18.117.158.108da2