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恐懼感8964 copyright protection364ＰＥＮＡＮＡm3TBphPdFj 維尼

368Please respect copyright.ＰＥＮＡＮＡZd7Tk8J997
8964 copyright protection364ＰＥＮＡＮＡ5aKHeH0DeX 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection364ＰＥＮＡＮＡ9EQPy8jCLJ 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection364ＰＥＮＡＮＡZ5eo2KQp97 維尼

=2∫eusin(u+a)du… or choose an alternative:368Please respect copyright.ＰＥＮＡＮＡfkoVekYpjX
Substitute e√x8964 copyright protection364ＰＥＮＡＮＡgCCFFZyw4o 維尼

Now solving:8964 copyright protection364ＰＥＮＡＮＡL0De4qE0aP 維尼

∫eusin(u+a)du8964 copyright protection364ＰＥＮＡＮＡfCNgIGrclT 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection364ＰＥＮＡＮＡ2ZBIgCR6lb 維尼

First time:8964 copyright protection364ＰＥＮＡＮＡtNLyl9BFDT 維尼

f=sin(u+a),g′=eu8964 copyright protection364ＰＥＮＡＮＡ6fVJOpCcOe 維尼

↓ steps↓ steps8964 copyright protection364ＰＥＮＡＮＡrc0spxC1Bv 維尼

f′=cos(u+a),g=eu:8964 copyright protection364ＰＥＮＡＮＡwgWdQ1hwFk 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection364ＰＥＮＡＮＡTB1YSSAY8D 維尼

Second time:8964 copyright protection364ＰＥＮＡＮＡidcBgEvNpe 維尼

f=cos(u+a),g′=eu8964 copyright protection364ＰＥＮＡＮＡpVVBucwHdB 維尼

↓ steps↓ steps8964 copyright protection364ＰＥＮＡＮＡ2tSy3EbrMU 維尼

f′=−sin(u+a),g=eu:8964 copyright protection364ＰＥＮＡＮＡv0sSN1SW9c 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection364ＰＥＮＡＮＡtyFV0ZkEbf 維尼

Apply linearity:8964 copyright protection364ＰＥＮＡＮＡhqTsycsfrG 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection364ＰＥＮＡＮＡCbfTFls3zr 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection364ＰＥＮＡＮＡhfvCnYpehb 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection364ＰＥＮＡＮＡZShBoyc0kG 維尼

Plug in solved integrals:8964 copyright protection364ＰＥＮＡＮＡruHWKzjAoA 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection364ＰＥＮＡＮＡiwWwiVG9i4 維尼

Undo substitution u=√x:8964 copyright protection364ＰＥＮＡＮＡLv94m0dgsh 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection364ＰＥＮＡＮＡ8a1JIZecaE 維尼

The problem is solved:8964 copyright protection364ＰＥＮＡＮＡoHcQsC3b9G 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection364ＰＥＮＡＮＡUcdurEo0su 維尼

Rewrite/simplify:8964 copyright protection364ＰＥＮＡＮＡvl4ZTT6pAu 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection364ＰＥＮＡＮＡCM7TEbnpXQ 維尼

15.158.61.12