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No Plagiarism!IQRoN8Ev9pM9B2DEZVjtposted on PＥNAＮＡ 恐懼感8964 copyright protection529ＰＥＮＡＮＡgteqZcvHDM 維尼
533Please respect copyright.ＰＥＮＡＮＡK0JvJcy4dQ
8964 copyright protection529ＰＥＮＡＮＡsqNaObMhsk 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection529ＰＥＮＡＮＡVx4kGPgwR3 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection529ＰＥＮＡＮＡJtddMS8YsT 維尼
=2∫eusin(u+a)du… or choose an alternative:533Please respect copyright.ＰＥＮＡＮＡkf2Ae36n9B
Substitute e√x8964 copyright protection529ＰＥＮＡＮＡPi0usre5gs 維尼
Now solving:8964 copyright protection529ＰＥＮＡＮＡP155GYO5Wc 維尼
∫eusin(u+a)du8964 copyright protection529ＰＥＮＡＮＡPGqrCmbIrW 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection529ＰＥＮＡＮＡRgLv4M19KY 維尼
First time:8964 copyright protection529ＰＥＮＡＮＡ2hTI4erk7n 維尼
f=sin(u+a),g′=eu8964 copyright protection529ＰＥＮＡＮＡT8jcvnYbmK 維尼
↓ steps↓ steps8964 copyright protection529ＰＥＮＡＮＡrr91hFxeWV 維尼
f′=cos(u+a),g=eu:8964 copyright protection529ＰＥＮＡＮＡB9XZjPQZfY 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection529ＰＥＮＡＮＡVYacxtPgms 維尼
Second time:8964 copyright protection529ＰＥＮＡＮＡfJqNFGwja7 維尼
f=cos(u+a),g′=eu8964 copyright protection529ＰＥＮＡＮＡw5cxGAXsVS 維尼
↓ steps↓ steps8964 copyright protection529ＰＥＮＡＮＡXhe5Uorfne 維尼
f′=−sin(u+a),g=eu:8964 copyright protection529ＰＥＮＡＮＡUC0MitWAse 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection529ＰＥＮＡＮＡuf3Sy28mZb 維尼
Apply linearity:8964 copyright protection529ＰＥＮＡＮＡGIXdxs3yAK 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection529ＰＥＮＡＮＡxpU8zAzi5k 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection529ＰＥＮＡＮＡHlDprp0Od4 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection529ＰＥＮＡＮＡJ7ROAGZdSh 維尼
Plug in solved integrals:8964 copyright protection529ＰＥＮＡＮＡVcL5lovwJ2 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection529ＰＥＮＡＮＡKF6U6izT2c 維尼
Undo substitution u=√x:8964 copyright protection529ＰＥＮＡＮＡI03ZstpFoD 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection529ＰＥＮＡＮＡZKdV0g7AN3 維尼
The problem is solved:8964 copyright protection529ＰＥＮＡＮＡYxNJfaMdLe 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection529ＰＥＮＡＮＡ0vknDlPQAG 維尼
Rewrite/simplify:8964 copyright protection529ＰＥＮＡＮＡCfpiswl0ky 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection529ＰＥＮＡＮＡUAI9mGnIwS 維尼
216.73.216.176
ns216.73.216.176da2
