No Plagiarism!0GrNd4oXnMG0QFNWf3Xuposted on PENANA 恐懼感8964 copyright protection301PENANAJ3wDzNaicw 維尼
305Please respect copyright.PENANAi2dd5kkFpp
8964 copyright protection301PENANAMs3H9p7cWu 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection301PENANAAY4guMxdHb 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection301PENANA6jyCJVIaLm 維尼
=2∫eusin(u+a)du… or choose an alternative:305Please respect copyright.PENANAeieSDjvWif
Substitute e√x8964 copyright protection301PENANABImkAWPK9L 維尼
Now solving:8964 copyright protection301PENANAYOhv0Z6cFL 維尼
∫eusin(u+a)du8964 copyright protection301PENANA6nDNQyAtjt 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection301PENANASf2o7oRzfN 維尼
First time:8964 copyright protection301PENANA8fa7HXfElj 維尼
f=sin(u+a),g′=eu8964 copyright protection301PENANAafJjYIvRgR 維尼
↓ steps↓ steps8964 copyright protection301PENANAp6WPEEG6pR 維尼
f′=cos(u+a),g=eu:8964 copyright protection301PENANA5Iklz9ZmiR 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection301PENANAcQRfEytwbU 維尼
Second time:8964 copyright protection301PENANAe08HHXgLXL 維尼
f=cos(u+a),g′=eu8964 copyright protection301PENANAWjf1lwuJ4A 維尼
↓ steps↓ steps8964 copyright protection301PENANAh75ORlocre 維尼
f′=−sin(u+a),g=eu:8964 copyright protection301PENANAfMIjr3l3k3 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection301PENANAVkZNby2Waw 維尼
Apply linearity:8964 copyright protection301PENANABibdJXCoJw 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection301PENANA7cEhQaufr9 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection301PENANAbA6g50Axnq 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection301PENANAA2Q9qVrH35 維尼
Plug in solved integrals:8964 copyright protection301PENANAut4dDBdB3u 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection301PENANAASWFG4APtG 維尼
Undo substitution u=√x:8964 copyright protection301PENANAuLh3E4foRr 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection301PENANAgWveAY1zdt 維尼
The problem is solved:8964 copyright protection301PENANAzeg3mvPKT5 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection301PENANAJwkCM8WoYY 維尼
Rewrite/simplify:8964 copyright protection301PENANAg2NFnjZj9z 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection301PENANA06BPQ71oEB 維尼
172.69.59.99
ns 172.69.59.99da2