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恐懼感8964 copyright protection550ＰＥＮＡＮＡMatuuL345i 維尼

554Please respect copyright.ＰＥＮＡＮＡ7qMRPf9aET
8964 copyright protection550ＰＥＮＡＮＡuhPrazUQIg 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection550ＰＥＮＡＮＡC6b0gbGLZB 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection550ＰＥＮＡＮＡ1Qn8Uqga0m 維尼

=2∫eusin(u+a)du… or choose an alternative:554Please respect copyright.ＰＥＮＡＮＡm7rgyZT4VV
Substitute e√x8964 copyright protection550ＰＥＮＡＮＡ2m7GxCg44b 維尼

Now solving:8964 copyright protection550ＰＥＮＡＮＡdWmPztuhZC 維尼

∫eusin(u+a)du8964 copyright protection550ＰＥＮＡＮＡWdPGegsZuk 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection550ＰＥＮＡＮＡZd8ZCPFYkG 維尼

First time:8964 copyright protection550ＰＥＮＡＮＡyUdkDYSqhX 維尼

f=sin(u+a),g′=eu8964 copyright protection550ＰＥＮＡＮＡNP7jihTOuH 維尼

↓ steps↓ steps8964 copyright protection550ＰＥＮＡＮＡw1RtldvkUU 維尼

f′=cos(u+a),g=eu:8964 copyright protection550ＰＥＮＡＮＡMVwL087lvj 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection550ＰＥＮＡＮＡ2fRF4GNJF3 維尼

Second time:8964 copyright protection550ＰＥＮＡＮＡcbbRt15jGY 維尼

f=cos(u+a),g′=eu8964 copyright protection550ＰＥＮＡＮＡwYxGrz515g 維尼

↓ steps↓ steps8964 copyright protection550ＰＥＮＡＮＡIwkoJvKJtC 維尼

f′=−sin(u+a),g=eu:8964 copyright protection550ＰＥＮＡＮＡUat7H23gct 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection550ＰＥＮＡＮＡPvJQz3BHjs 維尼

Apply linearity:8964 copyright protection550ＰＥＮＡＮＡoEDK3tqQaa 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection550ＰＥＮＡＮＡfVap3WxmBT 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection550ＰＥＮＡＮＡdvFLjyZ5Lz 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection550ＰＥＮＡＮＡwk349YJPPf 維尼

Plug in solved integrals:8964 copyright protection550ＰＥＮＡＮＡP4pzLRTRUx 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection550ＰＥＮＡＮＡIK4ZUHM3nP 維尼

Undo substitution u=√x:8964 copyright protection550ＰＥＮＡＮＡceDC2N3k5h 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection550ＰＥＮＡＮＡeAT5NRp7bm 維尼

The problem is solved:8964 copyright protection550ＰＥＮＡＮＡUQ93J8nwmA 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection550ＰＥＮＡＮＡniWfBHOuiE 維尼

Rewrite/simplify:8964 copyright protection550ＰＥＮＡＮＡZ91TAAfhfJ 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection550ＰＥＮＡＮＡ94WB9uQLld 維尼

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