x

恐懼感8964 copyright protection514ＰＥＮＡＮＡ6cnYEmdoGb 維尼

518Please respect copyright.ＰＥＮＡＮＡyiSBIRQeli
8964 copyright protection514ＰＥＮＡＮＡO2WvDUp2l5 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection514ＰＥＮＡＮＡYvCtvFJ3Sg 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection514ＰＥＮＡＮＡOFKQslPXTi 維尼

=2∫eusin(u+a)du… or choose an alternative:518Please respect copyright.ＰＥＮＡＮＡshDHuRIX5T
Substitute e√x8964 copyright protection514ＰＥＮＡＮＡEvmFyt44WD 維尼

Now solving:8964 copyright protection514ＰＥＮＡＮＡhmXmEJ64kY 維尼

∫eusin(u+a)du8964 copyright protection514ＰＥＮＡＮＡLdLQGYf6Il 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection514ＰＥＮＡＮＡfrfWcW5mBd 維尼

First time:8964 copyright protection514ＰＥＮＡＮＡeXi2NpOiND 維尼

f=sin(u+a),g′=eu8964 copyright protection514ＰＥＮＡＮＡXqfMuVrSy6 維尼

↓ steps↓ steps8964 copyright protection514ＰＥＮＡＮＡuBIOVSqxKL 維尼

f′=cos(u+a),g=eu:8964 copyright protection514ＰＥＮＡＮＡznDnBqPMSy 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection514ＰＥＮＡＮＡg0fKIWcJRn 維尼

Second time:8964 copyright protection514ＰＥＮＡＮＡBUusPBmWHP 維尼

f=cos(u+a),g′=eu8964 copyright protection514ＰＥＮＡＮＡv7062AfI9V 維尼

↓ steps↓ steps8964 copyright protection514ＰＥＮＡＮＡAJCPAfgG5X 維尼

f′=−sin(u+a),g=eu:8964 copyright protection514ＰＥＮＡＮＡ14SIx888Fc 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection514ＰＥＮＡＮＡIQhLJdycM5 維尼

Apply linearity:8964 copyright protection514ＰＥＮＡＮＡVYNneHlG51 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection514ＰＥＮＡＮＡDB9lrVe4eG 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection514ＰＥＮＡＮＡ45gTV791kv 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection514ＰＥＮＡＮＡwxen61sSY3 維尼

Plug in solved integrals:8964 copyright protection514ＰＥＮＡＮＡnkpUvLNaWh 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection514ＰＥＮＡＮＡAWKS9tnHC0 維尼

Undo substitution u=√x:8964 copyright protection514ＰＥＮＡＮＡM9kSoIa9V0 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection514ＰＥＮＡＮＡUiZ1k2chjE 維尼

The problem is solved:8964 copyright protection514ＰＥＮＡＮＡepiCSG06bg 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection514ＰＥＮＡＮＡ3FBckYZqJ1 維尼

Rewrite/simplify:8964 copyright protection514ＰＥＮＡＮＡbzsPXIRF5V 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection514ＰＥＮＡＮＡpxIaEdYYtt 維尼

216.73.216.0