No Plagiarism!3cn012H6RUmNXcvoZcPJposted on PENANA 恐懼感8964 copyright protection547PENANA7ElOPkhDKN 維尼
551Please respect copyright.PENANAbCTCSbvCIl
8964 copyright protection547PENANAwVnWyzS2Cz 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection547PENANA1souICxjSA 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection547PENANAzV3lepKGHd 維尼
=2∫eusin(u+a)du… or choose an alternative:551Please respect copyright.PENANAHzYB8kWGQf
Substitute e√x8964 copyright protection547PENANAElgfrlOtW5 維尼
Now solving:8964 copyright protection547PENANAKhjUUJeN28 維尼
∫eusin(u+a)du8964 copyright protection547PENANAY1roQmyhPD 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection547PENANAQ0fQzWhfgS 維尼
First time:8964 copyright protection547PENANAYp30zpSC7r 維尼
f=sin(u+a),g′=eu8964 copyright protection547PENANArCLPvHgFbf 維尼
↓ steps↓ steps8964 copyright protection547PENANA1YQaLGeVNy 維尼
f′=cos(u+a),g=eu:8964 copyright protection547PENANAS0ssGN06CJ 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection547PENANASToy0BEIbH 維尼
Second time:8964 copyright protection547PENANAnyHsSyJ4dN 維尼
f=cos(u+a),g′=eu8964 copyright protection547PENANAZxYCUZ4HmC 維尼
↓ steps↓ steps8964 copyright protection547PENANARn5eQqT5e0 維尼
f′=−sin(u+a),g=eu:8964 copyright protection547PENANAdstWqDbBHX 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection547PENANAaOemJq5Iui 維尼
Apply linearity:8964 copyright protection547PENANAnTXrvy5cEk 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection547PENANA2NlkjUr6gk 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection547PENANAXBIzZuljGW 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection547PENANAkdkCXCQG1O 維尼
Plug in solved integrals:8964 copyright protection547PENANADnNJz8G6N0 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection547PENANAjR1PJ6rpHr 維尼
Undo substitution u=√x:8964 copyright protection547PENANA4uprjZdkoQ 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection547PENANAs6AvPMLKrk 維尼
The problem is solved:8964 copyright protection547PENANAwuhTr4J9Hd 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection547PENANAawmatvNgnQ 維尼
Rewrite/simplify:8964 copyright protection547PENANA6SGp8yWmZF 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection547PENANAbcA70pKN4j 維尼
216.73.216.151
ns216.73.216.151da2