No Plagiarism!kxOWqkWfU92ZOCtbuu97posted on PENANA 恐懼感8964 copyright protection496PENANAAdEpaDoUox 維尼
500Please respect copyright.PENANASA5HZ8UwJe
8964 copyright protection496PENANAfXYEb9Czrz 維尼
∫sin(√x+a)e√x√xdx8964 copyright protection496PENANAmDdCJ3BBK2 維尼
Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection496PENANAn1TRpcG3Ya 維尼
=2∫eusin(u+a)du… or choose an alternative:500Please respect copyright.PENANApN6t1UicRd
Substitute e√x8964 copyright protection496PENANAsVCUbv017j 維尼
Now solving:8964 copyright protection496PENANA18aXUq8q33 維尼
∫eusin(u+a)du8964 copyright protection496PENANApBL89Yxu7z 維尼
We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection496PENANAVJbly1J88n 維尼
First time:8964 copyright protection496PENANAPm8J1e5snY 維尼
f=sin(u+a),g′=eu8964 copyright protection496PENANAnij6H90BM1 維尼
↓ steps↓ steps8964 copyright protection496PENANAcr5entixpU 維尼
f′=cos(u+a),g=eu:8964 copyright protection496PENANA8gHQJYCko8 維尼
=eusin(u+a)−∫eucos(u+a)du8964 copyright protection496PENANA9KZNI2Lsdk 維尼
Second time:8964 copyright protection496PENANAYT3L7nCMTi 維尼
f=cos(u+a),g′=eu8964 copyright protection496PENANAg25xiQHRx4 維尼
↓ steps↓ steps8964 copyright protection496PENANAMLUJjPmuex 維尼
f′=−sin(u+a),g=eu:8964 copyright protection496PENANAO3wH5ZhcR6 維尼
=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection496PENANAlz1YERIlC8 維尼
Apply linearity:8964 copyright protection496PENANAg8fEcLtOhb 維尼
=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection496PENANAqNgSkoUrGP 維尼
The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection496PENANAvyCnuleSlX 維尼
=eusin(u+a)−eucos(u+a)28964 copyright protection496PENANAsfz7QDVn2r 維尼
Plug in solved integrals:8964 copyright protection496PENANASaedM7Jkcm 維尼
2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection496PENANAtdJSo4maMq 維尼
Undo substitution u=√x:8964 copyright protection496PENANAGsFoDh4hTm 維尼
=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection496PENANA5yJrc9J0GI 維尼
The problem is solved:8964 copyright protection496PENANABYEnNIusLu 維尼
∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection496PENANAJJpuz7vmMO 維尼
Rewrite/simplify:8964 copyright protection496PENANAhgU4lkid1p 維尼
=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection496PENANA0WWcRZG4JZ 維尼
216.73.216.72
ns216.73.216.72da2