x

恐懼感8964 copyright protection417ＰＥＮＡＮＡVNMo6E0NYh 維尼

421Please respect copyright.ＰＥＮＡＮＡ65C6vdrONt
8964 copyright protection417ＰＥＮＡＮＡp2k0Vb25wy 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection417ＰＥＮＡＮＡ5aUIzyFP7B 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection417ＰＥＮＡＮＡxAlUbRHL3p 維尼

=2∫eusin(u+a)du… or choose an alternative:421Please respect copyright.ＰＥＮＡＮＡSi52fx6JfE
Substitute e√x8964 copyright protection417ＰＥＮＡＮＡGQA8Zlwwcq 維尼

Now solving:8964 copyright protection417ＰＥＮＡＮＡBwfwliGK9A 維尼

∫eusin(u+a)du8964 copyright protection417ＰＥＮＡＮＡR78rDhVaT0 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection417ＰＥＮＡＮＡYvSUgngtW6 維尼

First time:8964 copyright protection417ＰＥＮＡＮＡ8uAGrbmD9g 維尼

f=sin(u+a),g′=eu8964 copyright protection417ＰＥＮＡＮＡd9tk1M0vmQ 維尼

↓ steps↓ steps8964 copyright protection417ＰＥＮＡＮＡ6sAUtie0Gw 維尼

f′=cos(u+a),g=eu:8964 copyright protection417ＰＥＮＡＮＡZhdl5S5rld 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection417ＰＥＮＡＮＡ3Qr33rOJmA 維尼

Second time:8964 copyright protection417ＰＥＮＡＮＡ5MKQXH4F9E 維尼

f=cos(u+a),g′=eu8964 copyright protection417ＰＥＮＡＮＡXzBBht8Aet 維尼

↓ steps↓ steps8964 copyright protection417ＰＥＮＡＮＡQvGf7Ec6G5 維尼

f′=−sin(u+a),g=eu:8964 copyright protection417ＰＥＮＡＮＡ66Cf3gidyc 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection417ＰＥＮＡＮＡXnygZo5ZWg 維尼

Apply linearity:8964 copyright protection417ＰＥＮＡＮＡSCaPx4duV0 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection417ＰＥＮＡＮＡ6s57P0RAxV 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection417ＰＥＮＡＮＡoY5haEQbO4 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection417ＰＥＮＡＮＡFPA9cxUZHH 維尼

Plug in solved integrals:8964 copyright protection417ＰＥＮＡＮＡzeOD8MCRIK 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection417ＰＥＮＡＮＡbNdgcZMPUl 維尼

Undo substitution u=√x:8964 copyright protection417ＰＥＮＡＮＡHGD5Yy69cr 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection417ＰＥＮＡＮＡGNYWe9beiS 維尼

The problem is solved:8964 copyright protection417ＰＥＮＡＮＡxanWiUp1AN 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection417ＰＥＮＡＮＡw1V4sXJj6G 維尼

Rewrite/simplify:8964 copyright protection417ＰＥＮＡＮＡuvb9JjE0Cf 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection417ＰＥＮＡＮＡjWYZGxrNoo 維尼

18.188.149.185