x

恐懼感8964 copyright protection365ＰＥＮＡＮＡzbzzsTqtw0 維尼

369Please respect copyright.ＰＥＮＡＮＡk6sBCMnOT2
8964 copyright protection365ＰＥＮＡＮＡNeiA3ctprD 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection365ＰＥＮＡＮＡkkzLFG2cSF 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection365ＰＥＮＡＮＡ3EG0rvw9wr 維尼

=2∫eusin(u+a)du… or choose an alternative:369Please respect copyright.ＰＥＮＡＮＡ2EoODq91Z2
Substitute e√x8964 copyright protection365ＰＥＮＡＮＡ9UgJgkdYNB 維尼

Now solving:8964 copyright protection365ＰＥＮＡＮＡXHnNHFUXQn 維尼

∫eusin(u+a)du8964 copyright protection365ＰＥＮＡＮＡjsKrsJgg5S 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection365ＰＥＮＡＮＡGsFpUXwYKD 維尼

First time:8964 copyright protection365ＰＥＮＡＮＡqDm0htlLSc 維尼

f=sin(u+a),g′=eu8964 copyright protection365ＰＥＮＡＮＡLKRzvejI6x 維尼

↓ steps↓ steps8964 copyright protection365ＰＥＮＡＮＡkQFhWJDrGP 維尼

f′=cos(u+a),g=eu:8964 copyright protection365ＰＥＮＡＮＡLYz8NcVvR7 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection365ＰＥＮＡＮＡySXiZRdNTe 維尼

Second time:8964 copyright protection365ＰＥＮＡＮＡh0lCC7mW1Y 維尼

f=cos(u+a),g′=eu8964 copyright protection365ＰＥＮＡＮＡwvosHDFNEB 維尼

↓ steps↓ steps8964 copyright protection365ＰＥＮＡＮＡTagMiXzBX4 維尼

f′=−sin(u+a),g=eu:8964 copyright protection365ＰＥＮＡＮＡatcgE3LrAL 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection365ＰＥＮＡＮＡY0UnQNU3IQ 維尼

Apply linearity:8964 copyright protection365ＰＥＮＡＮＡSYeDCWNRox 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection365ＰＥＮＡＮＡPKpm63nZ6j 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection365ＰＥＮＡＮＡKNPR8woFuE 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection365ＰＥＮＡＮＡ15PGsJ8lFx 維尼

Plug in solved integrals:8964 copyright protection365ＰＥＮＡＮＡ3c4JQOJhML 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection365ＰＥＮＡＮＡ5xqQpOSSOv 維尼

Undo substitution u=√x:8964 copyright protection365ＰＥＮＡＮＡsCCmFENt8K 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection365ＰＥＮＡＮＡt7RvUgYKRk 維尼

The problem is solved:8964 copyright protection365ＰＥＮＡＮＡlpAw2IGyD9 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection365ＰＥＮＡＮＡXqaOugxUym 維尼

Rewrite/simplify:8964 copyright protection365ＰＥＮＡＮＡLwydaY6Lfs 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection365ＰＥＮＡＮＡefLcWjbHpx 維尼

15.158.61.54