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恐懼感8964 copyright protection379ＰＥＮＡＮＡPQ5CnIDCsM 維尼

383Please respect copyright.ＰＥＮＡＮＡjvVVU4e3jE
8964 copyright protection379ＰＥＮＡＮＡpIRDwaDN1H 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection379ＰＥＮＡＮＡQGtEqArLxO 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection379ＰＥＮＡＮＡzrIp0mJEhC 維尼

=2∫eusin(u+a)du… or choose an alternative:383Please respect copyright.ＰＥＮＡＮＡIF9hapk1ZX
Substitute e√x8964 copyright protection379ＰＥＮＡＮＡlQaVzAFsz0 維尼

Now solving:8964 copyright protection379ＰＥＮＡＮＡcV5Wu4AuJV 維尼

∫eusin(u+a)du8964 copyright protection379ＰＥＮＡＮＡ5dKcWosqJe 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection379ＰＥＮＡＮＡNXuL1RCMrE 維尼

First time:8964 copyright protection379ＰＥＮＡＮＡ2KcTTvJW67 維尼

f=sin(u+a),g′=eu8964 copyright protection379ＰＥＮＡＮＡ3Umr4alYmg 維尼

↓ steps↓ steps8964 copyright protection379ＰＥＮＡＮＡV9JLlZA3NZ 維尼

f′=cos(u+a),g=eu:8964 copyright protection379ＰＥＮＡＮＡgxPoovjd9l 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection379ＰＥＮＡＮＡC9aLXevCLp 維尼

Second time:8964 copyright protection379ＰＥＮＡＮＡ7Bp8K9Eggc 維尼

f=cos(u+a),g′=eu8964 copyright protection379ＰＥＮＡＮＡJcIEIJHeJ4 維尼

↓ steps↓ steps8964 copyright protection379ＰＥＮＡＮＡJE5dhIRqNO 維尼

f′=−sin(u+a),g=eu:8964 copyright protection379ＰＥＮＡＮＡtgmXJhzcdI 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection379ＰＥＮＡＮＡN0aswE7k7P 維尼

Apply linearity:8964 copyright protection379ＰＥＮＡＮＡy5mQTC2cO3 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection379ＰＥＮＡＮＡAtkYIvxJgI 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection379ＰＥＮＡＮＡPkT5XolYBA 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection379ＰＥＮＡＮＡJ3NUJkhXx1 維尼

Plug in solved integrals:8964 copyright protection379ＰＥＮＡＮＡbG2whRplE8 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection379ＰＥＮＡＮＡoUIY3Tpqsk 維尼

Undo substitution u=√x:8964 copyright protection379ＰＥＮＡＮＡo256cBO7x9 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection379ＰＥＮＡＮＡcnGNfnhdC0 維尼

The problem is solved:8964 copyright protection379ＰＥＮＡＮＡdK6lURkq6E 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection379ＰＥＮＡＮＡAloKdKGEog 維尼

Rewrite/simplify:8964 copyright protection379ＰＥＮＡＮＡXS08t3qXcK 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection379ＰＥＮＡＮＡbQxmPoyXkg 維尼

15.158.61.17