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恐懼感8964 copyright protection424ＰＥＮＡＮＡo5RsQNTltd 維尼

428Please respect copyright.ＰＥＮＡＮＡBBbFcLiLH4
8964 copyright protection424ＰＥＮＡＮＡ5aOLvpZemP 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection424ＰＥＮＡＮＡJc0VWzXG6o 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection424ＰＥＮＡＮＡ6CJwVUvt1v 維尼

=2∫eusin(u+a)du… or choose an alternative:428Please respect copyright.ＰＥＮＡＮＡKzDehuO7jS
Substitute e√x8964 copyright protection424ＰＥＮＡＮＡhDVUgKLuEV 維尼

Now solving:8964 copyright protection424ＰＥＮＡＮＡwzmBU1Egej 維尼

∫eusin(u+a)du8964 copyright protection424ＰＥＮＡＮＡ2Qxc3Ux2Q2 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection424ＰＥＮＡＮＡlbpeZkPwvi 維尼

First time:8964 copyright protection424ＰＥＮＡＮＡSKZY2mn7js 維尼

f=sin(u+a),g′=eu8964 copyright protection424ＰＥＮＡＮＡFachWAs8Aj 維尼

↓ steps↓ steps8964 copyright protection424ＰＥＮＡＮＡqHUQ9k2ws2 維尼

f′=cos(u+a),g=eu:8964 copyright protection424ＰＥＮＡＮＡYQBfn97gl7 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection424ＰＥＮＡＮＡZOkVUjGbUX 維尼

Second time:8964 copyright protection424ＰＥＮＡＮＡnwp6QdlPY3 維尼

f=cos(u+a),g′=eu8964 copyright protection424ＰＥＮＡＮＡRmR9NWcsj5 維尼

↓ steps↓ steps8964 copyright protection424ＰＥＮＡＮＡddtfhPZOFE 維尼

f′=−sin(u+a),g=eu:8964 copyright protection424ＰＥＮＡＮＡPQyfBCycGL 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection424ＰＥＮＡＮＡziJl7vGhp0 維尼

Apply linearity:8964 copyright protection424ＰＥＮＡＮＡLvPBn0W6kX 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection424ＰＥＮＡＮＡv3si459Tnl 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection424ＰＥＮＡＮＡKX08KuJXED 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection424ＰＥＮＡＮＡxVA07dYuA1 維尼

Plug in solved integrals:8964 copyright protection424ＰＥＮＡＮＡV9ddBriRPK 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection424ＰＥＮＡＮＡlp0FakgEzZ 維尼

Undo substitution u=√x:8964 copyright protection424ＰＥＮＡＮＡrPNNDGL0QN 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection424ＰＥＮＡＮＡIsnUQlT8zN 維尼

The problem is solved:8964 copyright protection424ＰＥＮＡＮＡTyfDGueyho 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection424ＰＥＮＡＮＡQDUqpxJrgM 維尼

Rewrite/simplify:8964 copyright protection424ＰＥＮＡＮＡhF9gI4MEHe 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection424ＰＥＮＡＮＡztvKeMVtUt 維尼

3.17.9.170