《極短篇主題式創挑》─恐懼篇 - 恐懼感

恐懼感

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∫sin(√x+a)e√x√xdx

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:

=2∫eusin(u+a)du… or choose an alternative:534Please respect copyright.ＰＥＮＡＮＡ4IcfL7xvCH
Substitute e√x

Now solving:

∫eusin(u+a)du

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

First time:

f=sin(u+a),g′=eu

f′=cos(u+a),g=eu:

=eusin(u+a)−∫eucos(u+a)du

Second time:

f=cos(u+a),g′=eu

f′=−sin(u+a),g=eu:

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)

Apply linearity:

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)

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

=eusin(u+a)−eucos(u+a)2

Plug in solved integrals:

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)

Undo substitution u=√x:

=sin(√x+a)e√x−cos(√x+a)e√x

The problem is solved:

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C

Rewrite/simplify:

=(sin(√x+a)−cos(√x+a))e√x+C

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