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No Plagiarism!i1dLPpcVTVmxzIKcLgoHposted on PＥNAＮＡ 恐懼感8964 copyright protection526ＰＥＮＡＮＡzmXiAAHDlO 維尼
530Please respect copyright.ＰＥＮＡＮＡ7cFuz2vezw
8964 copyright protection526ＰＥＮＡＮＡqPheHQrUw5 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection526ＰＥＮＡＮＡH9qsBZyfCV 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection526ＰＥＮＡＮＡdvqbvtm0N2 維尼
=2∫eusin(u+a)du… or choose an alternative:530Please respect copyright.ＰＥＮＡＮＡobsUY5iyqV
Substitute e√x8964 copyright protection526ＰＥＮＡＮＡFcMjVzrx51 維尼
Now solving:8964 copyright protection526ＰＥＮＡＮＡK2As6Q6OYZ 維尼
∫eusin(u+a)du8964 copyright protection526ＰＥＮＡＮＡiIDoGWI6tS 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection526ＰＥＮＡＮＡRcsabwFv1t 維尼
First time:8964 copyright protection526ＰＥＮＡＮＡRJzpnWVxVI 維尼
f=sin(u+a),g′=eu8964 copyright protection526ＰＥＮＡＮＡRIKrJczzfm 維尼
↓ steps↓ steps8964 copyright protection526ＰＥＮＡＮＡnbQzBmPmM4 維尼
f′=cos(u+a),g=eu:8964 copyright protection526ＰＥＮＡＮＡEOi9jwvokW 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection526ＰＥＮＡＮＡdxfvPQyegP 維尼
Second time:8964 copyright protection526ＰＥＮＡＮＡyZRPX6up7s 維尼
f=cos(u+a),g′=eu8964 copyright protection526ＰＥＮＡＮＡh9A6HnsRAa 維尼
↓ steps↓ steps8964 copyright protection526ＰＥＮＡＮＡ0T84A0Fhn1 維尼
f′=−sin(u+a),g=eu:8964 copyright protection526ＰＥＮＡＮＡMEGpRW4077 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection526ＰＥＮＡＮＡha8OGOTh6I 維尼
Apply linearity:8964 copyright protection526ＰＥＮＡＮＡTMzSJdTya2 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection526ＰＥＮＡＮＡRAU7nnbjTc 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection526ＰＥＮＡＮＡgnwralg77a 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection526ＰＥＮＡＮＡZyTmDe0QvA 維尼
Plug in solved integrals:8964 copyright protection526ＰＥＮＡＮＡbvchpy4Uo6 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection526ＰＥＮＡＮＡe8veOfM99s 維尼
Undo substitution u=√x:8964 copyright protection526ＰＥＮＡＮＡxf2PqQm2ua 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection526ＰＥＮＡＮＡPtAVus0EiZ 維尼
The problem is solved:8964 copyright protection526ＰＥＮＡＮＡKL4M97qjRB 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection526ＰＥＮＡＮＡMJBbA3eNrj 維尼
Rewrite/simplify:8964 copyright protection526ＰＥＮＡＮＡAcd1xPXOPY 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection526ＰＥＮＡＮＡSRE06lSXIG 維尼
216.73.216.146
ns216.73.216.146da2
