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恐懼感8964 copyright protection416ＰＥＮＡＮＡNoxDBpg57i 維尼

420Please respect copyright.ＰＥＮＡＮＡNnDaBNwKqf
8964 copyright protection416ＰＥＮＡＮＡmvDiZZETJF 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection416ＰＥＮＡＮＡv4OxIZSthC 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection416ＰＥＮＡＮＡpacwlzWSIS 維尼

=2∫eusin(u+a)du… or choose an alternative:420Please respect copyright.ＰＥＮＡＮＡvwI1w2jDJk
Substitute e√x8964 copyright protection416ＰＥＮＡＮＡnZ2zr1cWDp 維尼

Now solving:8964 copyright protection416ＰＥＮＡＮＡds3zQPpnCM 維尼

∫eusin(u+a)du8964 copyright protection416ＰＥＮＡＮＡSltxgVmEpt 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection416ＰＥＮＡＮＡJmRQO5RLFz 維尼

First time:8964 copyright protection416ＰＥＮＡＮＡpY5wi4yMsC 維尼

f=sin(u+a),g′=eu8964 copyright protection416ＰＥＮＡＮＡKsQfalbEGu 維尼

↓ steps↓ steps8964 copyright protection416ＰＥＮＡＮＡxSsTO6KJXD 維尼

f′=cos(u+a),g=eu:8964 copyright protection416ＰＥＮＡＮＡIg0IbmCJxx 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection416ＰＥＮＡＮＡ2z2XsJ6z3N 維尼

Second time:8964 copyright protection416ＰＥＮＡＮＡmvlC7mi70q 維尼

f=cos(u+a),g′=eu8964 copyright protection416ＰＥＮＡＮＡilPS05I1q5 維尼

↓ steps↓ steps8964 copyright protection416ＰＥＮＡＮＡA5a3Tk6lpR 維尼

f′=−sin(u+a),g=eu:8964 copyright protection416ＰＥＮＡＮＡIEWqkatibE 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection416ＰＥＮＡＮＡgTYW8m7he6 維尼

Apply linearity:8964 copyright protection416ＰＥＮＡＮＡky8cpaw6Ti 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection416ＰＥＮＡＮＡfYuTRkJ4Q9 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection416ＰＥＮＡＮＡD6EHfYiNXi 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection416ＰＥＮＡＮＡXTFZwSOOL4 維尼

Plug in solved integrals:8964 copyright protection416ＰＥＮＡＮＡtt6CZsD46M 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection416ＰＥＮＡＮＡrYgCkVy6Rx 維尼

Undo substitution u=√x:8964 copyright protection416ＰＥＮＡＮＡt0vQpj3AEV 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection416ＰＥＮＡＮＡSBH4AGWGM2 維尼

The problem is solved:8964 copyright protection416ＰＥＮＡＮＡsS0i6sfhWi 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection416ＰＥＮＡＮＡa7mIMvcPgn 維尼

Rewrite/simplify:8964 copyright protection416ＰＥＮＡＮＡToeHRS3O6g 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection416ＰＥＮＡＮＡofGbiC7Pos 維尼

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