No Plagiarism!iRtLEHQQUIsIoHiCeRPbposted on PENANA 恐懼感8964 copyright protection302PENANA0bigy98WGM 維尼
306Please respect copyright.PENANA49DXRmxa2i
8964 copyright protection302PENANAMr8XeYi2qC 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection302PENANA1CYrNCYJkU 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection302PENANAYOm8Pb1JYH 維尼
=2∫eusin(u+a)du… or choose an alternative:306Please respect copyright.PENANAHjCnevHqof
Substitute e√x8964 copyright protection302PENANAwD93DIi0u5 維尼
Now solving:8964 copyright protection302PENANAMdZEKEyKy1 維尼
∫eusin(u+a)du8964 copyright protection302PENANAZ8yNHCAG1E 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection302PENANAAT5UaHYSHh 維尼
First time:8964 copyright protection302PENANA2EhXTuQvxi 維尼
f=sin(u+a),g′=eu8964 copyright protection302PENANATtL2PuF0IT 維尼
↓ steps↓ steps8964 copyright protection302PENANAv2nPSvbujS 維尼
f′=cos(u+a),g=eu:8964 copyright protection302PENANAdufwdNWw9f 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection302PENANA6GMOGlAUei 維尼
Second time:8964 copyright protection302PENANAqORZpuATbb 維尼
f=cos(u+a),g′=eu8964 copyright protection302PENANAoCcbBw4n6I 維尼
↓ steps↓ steps8964 copyright protection302PENANAVoqcrtir31 維尼
f′=−sin(u+a),g=eu:8964 copyright protection302PENANAwwAqrI6m3Z 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection302PENANAGfu9N2cGYJ 維尼
Apply linearity:8964 copyright protection302PENANAF1VdmD08VO 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection302PENANAtSRh3SCjEI 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection302PENANAE8ZMzOXq7m 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection302PENANAbB2MAMRU82 維尼
Plug in solved integrals:8964 copyright protection302PENANAN1F0XAwjnq 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection302PENANAVmLYUuiZe9 維尼
Undo substitution u=√x:8964 copyright protection302PENANA9TKrYyBvNw 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection302PENANAD5viUue0Em 維尼
The problem is solved:8964 copyright protection302PENANANUoG6cSWOn 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection302PENANA1evVdlC0Rj 維尼
Rewrite/simplify:8964 copyright protection302PENANAVYXKVC66Wh 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection302PENANAh3EOshbaJD 維尼
172.70.126.52
ns 172.70.126.52da2