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No Plagiarism!285jHcjuBj4UTvmQykOJposted on PENANA 恐懼感8964 copyright protection550PENANAMatuuL345i 維尼
554Please respect copyright.PENANA7qMRPf9aET
8964 copyright protection550PENANAuhPrazUQIg 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection550PENANAC6b0gbGLZB 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection550PENANA1Qn8Uqga0m 維尼
=2∫eusin(u+a)du… or choose an alternative:554Please respect copyright.PENANAm7rgyZT4VV
Substitute e√x8964 copyright protection550PENANA2m7GxCg44b 維尼
Now solving:8964 copyright protection550PENANAdWmPztuhZC 維尼
∫eusin(u+a)du8964 copyright protection550PENANAWdPGegsZuk 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection550PENANAZd8ZCPFYkG 維尼
First time:8964 copyright protection550PENANAyUdkDYSqhX 維尼
f=sin(u+a),g′=eu8964 copyright protection550PENANANP7jihTOuH 維尼
↓ steps↓ steps8964 copyright protection550PENANAw1RtldvkUU 維尼
f′=cos(u+a),g=eu:8964 copyright protection550PENANAMVwL087lvj 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection550PENANA2fRF4GNJF3 維尼
Second time:8964 copyright protection550PENANAcbbRt15jGY 維尼
f=cos(u+a),g′=eu8964 copyright protection550PENANAwYxGrz515g 維尼
↓ steps↓ steps8964 copyright protection550PENANAIwkoJvKJtC 維尼
f′=−sin(u+a),g=eu:8964 copyright protection550PENANAUat7H23gct 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection550PENANAPvJQz3BHjs 維尼
Apply linearity:8964 copyright protection550PENANAoEDK3tqQaa 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection550PENANAfVap3WxmBT 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection550PENANAdvFLjyZ5Lz 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection550PENANAwk349YJPPf 維尼
Plug in solved integrals:8964 copyright protection550PENANAP4pzLRTRUx 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection550PENANAIK4ZUHM3nP 維尼
Undo substitution u=√x:8964 copyright protection550PENANAceDC2N3k5h 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection550PENANAeAT5NRp7bm 維尼
The problem is solved:8964 copyright protection550PENANAUQ93J8nwmA 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection550PENANAniWfBHOuiE 維尼
Rewrite/simplify:8964 copyright protection550PENANAZ91TAAfhfJ 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection550PENANA94WB9uQLld 維尼
216.73.216.25
ns216.73.216.25da2