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No Plagiarism!DLMb026oNWrirPzrJxmFposted on PＥNAＮＡ 恐懼感8964 copyright protection419ＰＥＮＡＮＡrOEmIxcFWY 維尼
423Please respect copyright.ＰＥＮＡＮＡaYtnfEmprk
8964 copyright protection419ＰＥＮＡＮＡdUuPPOyZbr 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection419ＰＥＮＡＮＡtWLEhjjBiM 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection419ＰＥＮＡＮＡ9p45gPEtBh 維尼
=2∫eusin(u+a)du… or choose an alternative:423Please respect copyright.ＰＥＮＡＮＡgllVZEcBhd
Substitute e√x8964 copyright protection419ＰＥＮＡＮＡy8SDndE7NG 維尼
Now solving:8964 copyright protection419ＰＥＮＡＮＡfkPkHkMPAi 維尼
∫eusin(u+a)du8964 copyright protection419ＰＥＮＡＮＡX8Gx69uqfe 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection419ＰＥＮＡＮＡZ2aNPoUUlG 維尼
First time:8964 copyright protection419ＰＥＮＡＮＡIC2nrhE8rw 維尼
f=sin(u+a),g′=eu8964 copyright protection419ＰＥＮＡＮＡLFcpM3Xa47 維尼
↓ steps↓ steps8964 copyright protection419ＰＥＮＡＮＡmuIY3iAA33 維尼
f′=cos(u+a),g=eu:8964 copyright protection419ＰＥＮＡＮＡVIDWBcJWc9 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection419ＰＥＮＡＮＡipkBeVf2c9 維尼
Second time:8964 copyright protection419ＰＥＮＡＮＡyyj7gZJJ5O 維尼
f=cos(u+a),g′=eu8964 copyright protection419ＰＥＮＡＮＡvT2SLRz1es 維尼
↓ steps↓ steps8964 copyright protection419ＰＥＮＡＮＡUo8kqNbD7y 維尼
f′=−sin(u+a),g=eu:8964 copyright protection419ＰＥＮＡＮＡJIka825XXA 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection419ＰＥＮＡＮＡuTr3oX81Xz 維尼
Apply linearity:8964 copyright protection419ＰＥＮＡＮＡDnVk5iQCBp 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection419ＰＥＮＡＮＡhBsvnHxDK4 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection419ＰＥＮＡＮＡ8v4R90oWME 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection419ＰＥＮＡＮＡtiCtKWB3bN 維尼
Plug in solved integrals:8964 copyright protection419ＰＥＮＡＮＡIhgMRnJ9vx 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection419ＰＥＮＡＮＡtOffi2gXtP 維尼
Undo substitution u=√x:8964 copyright protection419ＰＥＮＡＮＡAYBSq51Fa9 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection419ＰＥＮＡＮＡ2ASgp43WO2 維尼
The problem is solved:8964 copyright protection419ＰＥＮＡＮＡCNBL6ckydr 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection419ＰＥＮＡＮＡIigEzcqf6Y 維尼
Rewrite/simplify:8964 copyright protection419ＰＥＮＡＮＡqTxmH2Lkl0 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection419ＰＥＮＡＮＡ84QE9sbxDQ 維尼
18.188.189.204
ns18.188.189.204da2
