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恐懼感8964 copyright protection538ＰＥＮＡＮＡVMtcCpZu21 維尼

542Please respect copyright.ＰＥＮＡＮＡlGUERbAJ6G
8964 copyright protection538ＰＥＮＡＮＡbtltL4qq2s 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection538ＰＥＮＡＮＡxZMLV7d8RA 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection538ＰＥＮＡＮＡaoMsfKWpHB 維尼

=2∫eusin(u+a)du… or choose an alternative:542Please respect copyright.ＰＥＮＡＮＡgd20yn7r3t
Substitute e√x8964 copyright protection538ＰＥＮＡＮＡ6T17VZzOHj 維尼

Now solving:8964 copyright protection538ＰＥＮＡＮＡc81XINckFZ 維尼

∫eusin(u+a)du8964 copyright protection538ＰＥＮＡＮＡ8BjwtAw7Tq 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection538ＰＥＮＡＮＡcke2ZTc8Qt 維尼

First time:8964 copyright protection538ＰＥＮＡＮＡ8qkD4ld64T 維尼

f=sin(u+a),g′=eu8964 copyright protection538ＰＥＮＡＮＡGNHpEl7E4k 維尼

↓ steps↓ steps8964 copyright protection538ＰＥＮＡＮＡCly9qR7CtV 維尼

f′=cos(u+a),g=eu:8964 copyright protection538ＰＥＮＡＮＡtUDIwtuaET 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection538ＰＥＮＡＮＡeP2tmRcXU3 維尼

Second time:8964 copyright protection538ＰＥＮＡＮＡashqZg45sU 維尼

f=cos(u+a),g′=eu8964 copyright protection538ＰＥＮＡＮＡWG79hwGq99 維尼

↓ steps↓ steps8964 copyright protection538ＰＥＮＡＮＡR3y8IkQ6Bf 維尼

f′=−sin(u+a),g=eu:8964 copyright protection538ＰＥＮＡＮＡV7YjODIgCZ 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection538ＰＥＮＡＮＡKN6t4GUT7e 維尼

Apply linearity:8964 copyright protection538ＰＥＮＡＮＡCyq6hUNShn 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection538ＰＥＮＡＮＡgQO6Q52CLp 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection538ＰＥＮＡＮＡXSR2qIhduM 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection538ＰＥＮＡＮＡTjbg39Mz4i 維尼

Plug in solved integrals:8964 copyright protection538ＰＥＮＡＮＡMXKjDvh24A 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection538ＰＥＮＡＮＡLkBcXO4eqe 維尼

Undo substitution u=√x:8964 copyright protection538ＰＥＮＡＮＡvUB1QBtxR6 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection538ＰＥＮＡＮＡKQJhC7IVHM 維尼

The problem is solved:8964 copyright protection538ＰＥＮＡＮＡXq6GeGZuMD 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection538ＰＥＮＡＮＡDzD8SFHk5D 維尼

Rewrite/simplify:8964 copyright protection538ＰＥＮＡＮＡDOQVB4kxBh 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection538ＰＥＮＡＮＡ48vvzbd9Sm 維尼

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