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恐懼感8964 copyright protection398ＰＥＮＡＮＡHArNoURi2x 維尼

402Please respect copyright.ＰＥＮＡＮＡsnSUYE8llc
8964 copyright protection398ＰＥＮＡＮＡ79TOEbtlqL 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection398ＰＥＮＡＮＡ93s63EOK3H 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection398ＰＥＮＡＮＡ1yhrxwiTMr 維尼

=2∫eusin(u+a)du… or choose an alternative:402Please respect copyright.ＰＥＮＡＮＡq03YEzFqJS
Substitute e√x8964 copyright protection398ＰＥＮＡＮＡj5PsYE3afg 維尼

Now solving:8964 copyright protection398ＰＥＮＡＮＡTDcqoAii4p 維尼

∫eusin(u+a)du8964 copyright protection398ＰＥＮＡＮＡn8Kp0olO2E 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection398ＰＥＮＡＮＡbBgJGWx0qH 維尼

First time:8964 copyright protection398ＰＥＮＡＮＡo6QwvpxmR4 維尼

f=sin(u+a),g′=eu8964 copyright protection398ＰＥＮＡＮＡQWT1aTdDrs 維尼

↓ steps↓ steps8964 copyright protection398ＰＥＮＡＮＡ9rQla3LH3h 維尼

f′=cos(u+a),g=eu:8964 copyright protection398ＰＥＮＡＮＡBZXph0dAQy 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection398ＰＥＮＡＮＡlzIi4vNqkn 維尼

Second time:8964 copyright protection398ＰＥＮＡＮＡRvoPtrPyzC 維尼

f=cos(u+a),g′=eu8964 copyright protection398ＰＥＮＡＮＡyYn7acY98w 維尼

↓ steps↓ steps8964 copyright protection398ＰＥＮＡＮＡyzTVDcwGi2 維尼

f′=−sin(u+a),g=eu:8964 copyright protection398ＰＥＮＡＮＡ3NgEYfjtcC 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection398ＰＥＮＡＮＡxFD55pg4Tj 維尼

Apply linearity:8964 copyright protection398ＰＥＮＡＮＡoQo8jHEwLq 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection398ＰＥＮＡＮＡWeOqZSGtYV 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection398ＰＥＮＡＮＡTv0oSMQjev 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection398ＰＥＮＡＮＡCKnei7cWKy 維尼

Plug in solved integrals:8964 copyright protection398ＰＥＮＡＮＡtM1zNVAWpn 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection398ＰＥＮＡＮＡiie05FzkbB 維尼

Undo substitution u=√x:8964 copyright protection398ＰＥＮＡＮＡPhe0o9Dptu 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection398ＰＥＮＡＮＡP6zlOUw6Da 維尼

The problem is solved:8964 copyright protection398ＰＥＮＡＮＡ8dstJv58js 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection398ＰＥＮＡＮＡWkI2ACt4D7 維尼

Rewrite/simplify:8964 copyright protection398ＰＥＮＡＮＡbyfFxVstgK 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection398ＰＥＮＡＮＡATvdxtVUjo 維尼

15.158.61.4