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恐懼感8964 copyright protection546ＰＥＮＡＮＡl8uxYRDJUC 維尼

550Please respect copyright.ＰＥＮＡＮＡPUjLrdngW9
8964 copyright protection546ＰＥＮＡＮＡct1eJJyerk 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection546ＰＥＮＡＮＡS4FRC23Twx 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection546ＰＥＮＡＮＡuy5u4KWfDO 維尼

=2∫eusin(u+a)du… or choose an alternative:550Please respect copyright.ＰＥＮＡＮＡ2X1ImIofbS
Substitute e√x8964 copyright protection546ＰＥＮＡＮＡIZx4nirnNp 維尼

Now solving:8964 copyright protection546ＰＥＮＡＮＡS182hLRSVi 維尼

∫eusin(u+a)du8964 copyright protection546ＰＥＮＡＮＡuorPMKN6ZW 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection546ＰＥＮＡＮＡXGfxAlD1iB 維尼

First time:8964 copyright protection546ＰＥＮＡＮＡiLg5ygyoXh 維尼

f=sin(u+a),g′=eu8964 copyright protection546ＰＥＮＡＮＡ3kFItXSLoO 維尼

↓ steps↓ steps8964 copyright protection546ＰＥＮＡＮＡJJh0wxfleY 維尼

f′=cos(u+a),g=eu:8964 copyright protection546ＰＥＮＡＮＡ3vAdTe6HtW 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection546ＰＥＮＡＮＡqWNDUSTKU3 維尼

Second time:8964 copyright protection546ＰＥＮＡＮＡNjQEzm3jGE 維尼

f=cos(u+a),g′=eu8964 copyright protection546ＰＥＮＡＮＡ6E3n886PkY 維尼

↓ steps↓ steps8964 copyright protection546ＰＥＮＡＮＡhx32pxurVa 維尼

f′=−sin(u+a),g=eu:8964 copyright protection546ＰＥＮＡＮＡ89Rnxfh2La 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection546ＰＥＮＡＮＡsXzQL03R7v 維尼

Apply linearity:8964 copyright protection546ＰＥＮＡＮＡhsGZWvVJ3i 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection546ＰＥＮＡＮＡJsM6ksHIFW 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection546ＰＥＮＡＮＡdXWs3rd2Bn 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection546ＰＥＮＡＮＡyvYfGShVqU 維尼

Plug in solved integrals:8964 copyright protection546ＰＥＮＡＮＡpZEkZSPrhE 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection546ＰＥＮＡＮＡPDmV53v7uV 維尼

Undo substitution u=√x:8964 copyright protection546ＰＥＮＡＮＡVlIM4nKjpz 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection546ＰＥＮＡＮＡyiHY7l2OzJ 維尼

The problem is solved:8964 copyright protection546ＰＥＮＡＮＡqoJoCkbI4I 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection546ＰＥＮＡＮＡOkruEfGZHj 維尼

Rewrite/simplify:8964 copyright protection546ＰＥＮＡＮＡivZpDeX5Pr 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection546ＰＥＮＡＮＡ7sXCb1l29A 維尼

216.73.216.3