x

恐懼感8964 copyright protection423ＰＥＮＡＮＡ716jRx1I4C 維尼

427Please respect copyright.ＰＥＮＡＮＡL9mQj61TNP
8964 copyright protection423ＰＥＮＡＮＡjhoAkdID4C 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection423ＰＥＮＡＮＡzCgnwqITQl 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection423ＰＥＮＡＮＡWfgqQaoq8t 維尼

=2∫eusin(u+a)du… or choose an alternative:427Please respect copyright.ＰＥＮＡＮＡefjMsHpWRN
Substitute e√x8964 copyright protection423ＰＥＮＡＮＡiFshhub66N 維尼

Now solving:8964 copyright protection423ＰＥＮＡＮＡkQEoE0OSMF 維尼

∫eusin(u+a)du8964 copyright protection423ＰＥＮＡＮＡCMeB3aHBD0 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection423ＰＥＮＡＮＡhQYBrR5uEw 維尼

First time:8964 copyright protection423ＰＥＮＡＮＡBFBDI4HHPJ 維尼

f=sin(u+a),g′=eu8964 copyright protection423ＰＥＮＡＮＡX5dXOJIp9j 維尼

↓ steps↓ steps8964 copyright protection423ＰＥＮＡＮＡt5N1S2itpU 維尼

f′=cos(u+a),g=eu:8964 copyright protection423ＰＥＮＡＮＡXKAC7PZfTb 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection423ＰＥＮＡＮＡbYUz1kakHY 維尼

Second time:8964 copyright protection423ＰＥＮＡＮＡv3NriZGipx 維尼

f=cos(u+a),g′=eu8964 copyright protection423ＰＥＮＡＮＡi4nnQnYDKo 維尼

↓ steps↓ steps8964 copyright protection423ＰＥＮＡＮＡepzrXxfsOx 維尼

f′=−sin(u+a),g=eu:8964 copyright protection423ＰＥＮＡＮＡHfpFeD90cP 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection423ＰＥＮＡＮＡttczmJRtZF 維尼

Apply linearity:8964 copyright protection423ＰＥＮＡＮＡx6RM59lc6N 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection423ＰＥＮＡＮＡ291ndn1DJk 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection423ＰＥＮＡＮＡgIRXSr6Yji 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection423ＰＥＮＡＮＡQc3CjQu1jr 維尼

Plug in solved integrals:8964 copyright protection423ＰＥＮＡＮＡ5yEfvJOUp0 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection423ＰＥＮＡＮＡXvosPeAvcL 維尼

Undo substitution u=√x:8964 copyright protection423ＰＥＮＡＮＡrHCX1EmKUQ 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection423ＰＥＮＡＮＡ7Du03KqyTw 維尼

The problem is solved:8964 copyright protection423ＰＥＮＡＮＡqmXU3Q0al7 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection423ＰＥＮＡＮＡ9uc1dhy3Pk 維尼

Rewrite/simplify:8964 copyright protection423ＰＥＮＡＮＡLPnPo0v2H8 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection423ＰＥＮＡＮＡqCscwzi6gA 維尼

3.19.64.3