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恐懼感8964 copyright protection535ＰＥＮＡＮＡmhCnEYN3f1 維尼

539Please respect copyright.ＰＥＮＡＮＡUBDrf95QzO
8964 copyright protection535ＰＥＮＡＮＡEWmjXs3ho6 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection535ＰＥＮＡＮＡO8pmxJGe6p 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection535ＰＥＮＡＮＡC54YdVI13k 維尼

=2∫eusin(u+a)du… or choose an alternative:539Please respect copyright.ＰＥＮＡＮＡ3hZNo3yu1d
Substitute e√x8964 copyright protection535ＰＥＮＡＮＡXwOUThY8z1 維尼

Now solving:8964 copyright protection535ＰＥＮＡＮＡC7Eq7hmBl7 維尼

∫eusin(u+a)du8964 copyright protection535ＰＥＮＡＮＡI4cnmnpyW9 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection535ＰＥＮＡＮＡefzo2OpnsN 維尼

First time:8964 copyright protection535ＰＥＮＡＮＡIwMK2MWrww 維尼

f=sin(u+a),g′=eu8964 copyright protection535ＰＥＮＡＮＡcpjcvo1yOM 維尼

↓ steps↓ steps8964 copyright protection535ＰＥＮＡＮＡiHoF6iL5gl 維尼

f′=cos(u+a),g=eu:8964 copyright protection535ＰＥＮＡＮＡV0hnX6v23Q 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection535ＰＥＮＡＮＡCt403sxmkR 維尼

Second time:8964 copyright protection535ＰＥＮＡＮＡP6u5OTYDdN 維尼

f=cos(u+a),g′=eu8964 copyright protection535ＰＥＮＡＮＡZpS0QWqTy2 維尼

↓ steps↓ steps8964 copyright protection535ＰＥＮＡＮＡ40IY9UYtK8 維尼

f′=−sin(u+a),g=eu:8964 copyright protection535ＰＥＮＡＮＡyNbe1tzHQd 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection535ＰＥＮＡＮＡvI8mDGEnie 維尼

Apply linearity:8964 copyright protection535ＰＥＮＡＮＡPYL8anvsmO 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection535ＰＥＮＡＮＡDcb0ujm9ki 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection535ＰＥＮＡＮＡXJQkgQN9lb 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection535ＰＥＮＡＮＡrugN6XaH6Z 維尼

Plug in solved integrals:8964 copyright protection535ＰＥＮＡＮＡ23GbZ7hq58 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection535ＰＥＮＡＮＡ7UDxAgp3kM 維尼

Undo substitution u=√x:8964 copyright protection535ＰＥＮＡＮＡuXtf0vzjch 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection535ＰＥＮＡＮＡBv9SHuiP4i 維尼

The problem is solved:8964 copyright protection535ＰＥＮＡＮＡwr7wkkTewT 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection535ＰＥＮＡＮＡYVcT4X230N 維尼

Rewrite/simplify:8964 copyright protection535ＰＥＮＡＮＡxeNjX1pwNJ 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection535ＰＥＮＡＮＡkROjf6baS9 維尼

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