x

恐懼感8964 copyright protection550ＰＥＮＡＮＡtLYaSxE2Ze 維尼

554Please respect copyright.ＰＥＮＡＮＡ3mUaqqcr9P
8964 copyright protection550ＰＥＮＡＮＡn0wma6lkiE 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection550ＰＥＮＡＮＡW2sbwGJOT6 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection550ＰＥＮＡＮＡHE31rzXb6N 維尼

=2∫eusin(u+a)du… or choose an alternative:554Please respect copyright.ＰＥＮＡＮＡ5w7PZv6JN5
Substitute e√x8964 copyright protection550ＰＥＮＡＮＡPraFcERjc1 維尼

Now solving:8964 copyright protection550ＰＥＮＡＮＡc9ly1dg2BJ 維尼

∫eusin(u+a)du8964 copyright protection550ＰＥＮＡＮＡSkSxkDli7X 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection550ＰＥＮＡＮＡwD3bW25H8n 維尼

First time:8964 copyright protection550ＰＥＮＡＮＡEKMazL398Z 維尼

f=sin(u+a),g′=eu8964 copyright protection550ＰＥＮＡＮＡLrxdmcAQRa 維尼

↓ steps↓ steps8964 copyright protection550ＰＥＮＡＮＡT2TQBfjZUx 維尼

f′=cos(u+a),g=eu:8964 copyright protection550ＰＥＮＡＮＡFUaf868p7X 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection550ＰＥＮＡＮＡIeon68en7m 維尼

Second time:8964 copyright protection550ＰＥＮＡＮＡgncNdwA8Cu 維尼

f=cos(u+a),g′=eu8964 copyright protection550ＰＥＮＡＮＡCNdOXcSByM 維尼

↓ steps↓ steps8964 copyright protection550ＰＥＮＡＮＡNwUSDAwVLQ 維尼

f′=−sin(u+a),g=eu:8964 copyright protection550ＰＥＮＡＮＡGXOyg1RxAx 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection550ＰＥＮＡＮＡVOeiKdORAn 維尼

Apply linearity:8964 copyright protection550ＰＥＮＡＮＡfKQLv6UIhU 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection550ＰＥＮＡＮＡXZEIGQjPtQ 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection550ＰＥＮＡＮＡm1tm8O62lN 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection550ＰＥＮＡＮＡJwJKW6lO7Y 維尼

Plug in solved integrals:8964 copyright protection550ＰＥＮＡＮＡ0NU3BPuC5c 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection550ＰＥＮＡＮＡLciB1pDhJ4 維尼

Undo substitution u=√x:8964 copyright protection550ＰＥＮＡＮＡ4xjKi8kT7D 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection550ＰＥＮＡＮＡyeDZnCi6cE 維尼

The problem is solved:8964 copyright protection550ＰＥＮＡＮＡeURzFVAhQ4 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection550ＰＥＮＡＮＡpSPSx9MWUU 維尼

Rewrite/simplify:8964 copyright protection550ＰＥＮＡＮＡKt8De6MzuT 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection550ＰＥＮＡＮＡoLMJAqlg4H 維尼

216.73.216.25