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No Plagiarism!B98VN8ugOdH7VYwYVNuuposted on PENANA 恐懼感8964 copyright protection312PENANAEZx9vXvQ4r 維尼
316Please respect copyright.PENANADXlJWD9fY9
8964 copyright protection312PENANAJ5pErdPTAu 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection312PENANAfJZg0cFzjh 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection312PENANANh099VGyRb 維尼
=2∫eusin(u+a)du… or choose an alternative:316Please respect copyright.PENANADxitJsIAfC
Substitute e√x8964 copyright protection312PENANANyVa1gclAo 維尼
Now solving:8964 copyright protection312PENANAt5sD4YnORz 維尼
∫eusin(u+a)du8964 copyright protection312PENANAVvxo3h4FNm 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection312PENANAwox6AKsD7C 維尼
First time:8964 copyright protection312PENANAIAtlM8Rhwi 維尼
f=sin(u+a),g′=eu8964 copyright protection312PENANADMJ258SZHV 維尼
↓ steps↓ steps8964 copyright protection312PENANAklnyBcmmta 維尼
f′=cos(u+a),g=eu:8964 copyright protection312PENANAX1X9efL71g 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection312PENANA2uupBwXz6g 維尼
Second time:8964 copyright protection312PENANAEuwhw1DjYE 維尼
f=cos(u+a),g′=eu8964 copyright protection312PENANAB62urM4JDO 維尼
↓ steps↓ steps8964 copyright protection312PENANAGzST64mAF7 維尼
f′=−sin(u+a),g=eu:8964 copyright protection312PENANAWheT8Dvw7K 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection312PENANAbWEjwnVy9f 維尼
Apply linearity:8964 copyright protection312PENANA2AT2fNs4fo 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection312PENANAWVhAB0iepM 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection312PENANAnZMgiIS2SR 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection312PENANAgyRYCaYrKE 維尼
Plug in solved integrals:8964 copyright protection312PENANADLRglW7HF4 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection312PENANAHMvNNkYfew 維尼
Undo substitution u=√x:8964 copyright protection312PENANAy0WMupSIcq 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection312PENANATh7VWV8Kcl 維尼
The problem is solved:8964 copyright protection312PENANAhFs6ZyFDV2 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection312PENANAvmAeWbO3yP 維尼
Rewrite/simplify:8964 copyright protection312PENANAcggAyAZkeH 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection312PENANAKwfA8DETOn 維尼
172.69.7.3
ns 172.69.7.3da2