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No Plagiarism!5zp9uzktBIA2p5Oc1Vjtposted on PENANA 恐懼感8964 copyright protection289PENANAMtsvAGz6ee 維尼
293Please respect copyright.PENANAU6T24G428v
8964 copyright protection289PENANAtEMBMNoRYw 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection289PENANAsgTKwaqHkw 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection289PENANACrzHTLaiFv 維尼
=2∫eusin(u+a)du… or choose an alternative:293Please respect copyright.PENANANqAZc91QLA
Substitute e√x8964 copyright protection289PENANARlCoAUHt5E 維尼
Now solving:8964 copyright protection289PENANAhGQIndTvzY 維尼
∫eusin(u+a)du8964 copyright protection289PENANAAThfxjDoVy 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection289PENANAEfQDgEio6k 維尼
First time:8964 copyright protection289PENANAGlvURIG1Lb 維尼
f=sin(u+a),g′=eu8964 copyright protection289PENANAo5kqInJNBs 維尼
↓ steps↓ steps8964 copyright protection289PENANA9fi2SOSlME 維尼
f′=cos(u+a),g=eu:8964 copyright protection289PENANA9VADbMTpgc 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection289PENANAx9tMvObFQV 維尼
Second time:8964 copyright protection289PENANA3UoPGseAhm 維尼
f=cos(u+a),g′=eu8964 copyright protection289PENANAarikRzXIh0 維尼
↓ steps↓ steps8964 copyright protection289PENANAMtYjCfdrz6 維尼
f′=−sin(u+a),g=eu:8964 copyright protection289PENANALvKBak7ebP 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection289PENANAXG6l6Lydzh 維尼
Apply linearity:8964 copyright protection289PENANA4vym01ZzXd 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection289PENANAhfUu97mAai 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection289PENANAAvuBM07h8W 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection289PENANAAPr4EdezAd 維尼
Plug in solved integrals:8964 copyright protection289PENANAImO8byK4N0 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection289PENANAcHBw1XPObJ 維尼
Undo substitution u=√x:8964 copyright protection289PENANAvnty7dHb4y 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection289PENANAANi1xV8bWa 維尼
The problem is solved:8964 copyright protection289PENANAMTNF0w4MeJ 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection289PENANAJMYx5xTHGL 維尼
Rewrite/simplify:8964 copyright protection289PENANAdjg1r1hi58 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection289PENANAFILHagw0EQ 維尼
172.69.59.183
ns 172.69.59.183da2