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No Plagiarism!3n9CNtOPcm5iGzdwGglaposted on PＥNAＮＡ 恐懼感8964 copyright protection451ＰＥＮＡＮＡTA3UFg9CFs 維尼
455Please respect copyright.ＰＥＮＡＮＡcA7vB6fHtu
8964 copyright protection451ＰＥＮＡＮＡm4aZQhSu9y 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection451ＰＥＮＡＮＡVgzDcmux2m 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection451ＰＥＮＡＮＡ07LGk8fK3G 維尼
=2∫eusin(u+a)du… or choose an alternative:455Please respect copyright.ＰＥＮＡＮＡKkxELQdmvf
Substitute e√x8964 copyright protection451ＰＥＮＡＮＡWUaR93pWtz 維尼
Now solving:8964 copyright protection451ＰＥＮＡＮＡy4Z2NEOhZT 維尼
∫eusin(u+a)du8964 copyright protection451ＰＥＮＡＮＡBzCOytPqjv 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection451ＰＥＮＡＮＡJ35SdGz3At 維尼
First time:8964 copyright protection451ＰＥＮＡＮＡnirIE6fb69 維尼
f=sin(u+a),g′=eu8964 copyright protection451ＰＥＮＡＮＡNasGihx7Q9 維尼
↓ steps↓ steps8964 copyright protection451ＰＥＮＡＮＡUo79ONfdh2 維尼
f′=cos(u+a),g=eu:8964 copyright protection451ＰＥＮＡＮＡnlnwTuw8aj 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection451ＰＥＮＡＮＡilS3lXZd7A 維尼
Second time:8964 copyright protection451ＰＥＮＡＮＡZNEGHgVdCr 維尼
f=cos(u+a),g′=eu8964 copyright protection451ＰＥＮＡＮＡZ6Y4yXPwMe 維尼
↓ steps↓ steps8964 copyright protection451ＰＥＮＡＮＡ6Cc6axHnPY 維尼
f′=−sin(u+a),g=eu:8964 copyright protection451ＰＥＮＡＮＡkFyxVvsFxL 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection451ＰＥＮＡＮＡ9PaVx1DHvr 維尼
Apply linearity:8964 copyright protection451ＰＥＮＡＮＡnztGfc5Bx7 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection451ＰＥＮＡＮＡ0adxvcP42O 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection451ＰＥＮＡＮＡ0fuafZiHNd 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection451ＰＥＮＡＮＡEvZi3rsoyM 維尼
Plug in solved integrals:8964 copyright protection451ＰＥＮＡＮＡkSjd6aLWgT 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection451ＰＥＮＡＮＡMFw1rFRuwI 維尼
Undo substitution u=√x:8964 copyright protection451ＰＥＮＡＮＡe79KXcQeWJ 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection451ＰＥＮＡＮＡq4xzwYdipD 維尼
The problem is solved:8964 copyright protection451ＰＥＮＡＮＡmXLoxjxu2O 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection451ＰＥＮＡＮＡwQlXsyVUzo 維尼
Rewrite/simplify:8964 copyright protection451ＰＥＮＡＮＡkv0H6KC6QD 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection451ＰＥＮＡＮＡM49Uq9OwoQ 維尼
3.146.206.173
ns3.146.206.173da2
