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恐懼感8964 copyright protection550ＰＥＮＡＮＡOWYsFqUYNW 維尼

554Please respect copyright.ＰＥＮＡＮＡS6wnRebsIv
8964 copyright protection550ＰＥＮＡＮＡV18P6SJjXk 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection550ＰＥＮＡＮＡCZ9vgreFKL 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection550ＰＥＮＡＮＡOk3p0Bk7gY 維尼

=2∫eusin(u+a)du… or choose an alternative:554Please respect copyright.ＰＥＮＡＮＡ4997ruLSG4
Substitute e√x8964 copyright protection550ＰＥＮＡＮＡ5yYa8wnUUn 維尼

Now solving:8964 copyright protection550ＰＥＮＡＮＡDhZoMoG5bn 維尼

∫eusin(u+a)du8964 copyright protection550ＰＥＮＡＮＡIKzYBrzUFU 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection550ＰＥＮＡＮＡ3eYj4A7Gt8 維尼

First time:8964 copyright protection550ＰＥＮＡＮＡdqnnBcgzE9 維尼

f=sin(u+a),g′=eu8964 copyright protection550ＰＥＮＡＮＡKk07CmoIdp 維尼

↓ steps↓ steps8964 copyright protection550ＰＥＮＡＮＡtYi6FAR6vi 維尼

f′=cos(u+a),g=eu:8964 copyright protection550ＰＥＮＡＮＡJPT637rMlA 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection550ＰＥＮＡＮＡA190fyvQUm 維尼

Second time:8964 copyright protection550ＰＥＮＡＮＡq6TKpCXjV3 維尼

f=cos(u+a),g′=eu8964 copyright protection550ＰＥＮＡＮＡmdpBvxXIdi 維尼

↓ steps↓ steps8964 copyright protection550ＰＥＮＡＮＡQ2ClEV4D4c 維尼

f′=−sin(u+a),g=eu:8964 copyright protection550ＰＥＮＡＮＡNwua6HYt8W 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection550ＰＥＮＡＮＡpG5UfRf2RT 維尼

Apply linearity:8964 copyright protection550ＰＥＮＡＮＡqZE5UDy93f 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection550ＰＥＮＡＮＡvS80SnUV5E 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection550ＰＥＮＡＮＡ0O7rFc840Y 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection550ＰＥＮＡＮＡKfedE6NEvi 維尼

Plug in solved integrals:8964 copyright protection550ＰＥＮＡＮＡODHguoK1U0 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection550ＰＥＮＡＮＡfnrk4GuRiR 維尼

Undo substitution u=√x:8964 copyright protection550ＰＥＮＡＮＡzthKJva0Of 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection550ＰＥＮＡＮＡ2KpTwVZFgI 維尼

The problem is solved:8964 copyright protection550ＰＥＮＡＮＡINMizIVArl 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection550ＰＥＮＡＮＡOFrzZsao1p 維尼

Rewrite/simplify:8964 copyright protection550ＰＥＮＡＮＡdTsElmho3p 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection550ＰＥＮＡＮＡH7l6RoQulH 維尼

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