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恐懼感8964 copyright protection537ＰＥＮＡＮＡb7VSTbZ9h1 維尼

541Please respect copyright.ＰＥＮＡＮＡ80tHzpVPpO
8964 copyright protection537ＰＥＮＡＮＡI1GHC8pXOu 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection537ＰＥＮＡＮＡ9gf9wmYT9x 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection537ＰＥＮＡＮＡBHqQ99N5Kc 維尼

=2∫eusin(u+a)du… or choose an alternative:541Please respect copyright.ＰＥＮＡＮＡS5hT6s7ZFv
Substitute e√x8964 copyright protection537ＰＥＮＡＮＡ9vBNcCUuTa 維尼

Now solving:8964 copyright protection537ＰＥＮＡＮＡmZjg2zQYGb 維尼

∫eusin(u+a)du8964 copyright protection537ＰＥＮＡＮＡAAfTmHSF5E 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection537ＰＥＮＡＮＡYch43ix7QE 維尼

First time:8964 copyright protection537ＰＥＮＡＮＡ0UxQmJev3E 維尼

f=sin(u+a),g′=eu8964 copyright protection537ＰＥＮＡＮＡ4zftAM1S7r 維尼

↓ steps↓ steps8964 copyright protection537ＰＥＮＡＮＡL1T3NU6Ux8 維尼

f′=cos(u+a),g=eu:8964 copyright protection537ＰＥＮＡＮＡ3gMglfhLh3 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection537ＰＥＮＡＮＡiIFS4Spmgd 維尼

Second time:8964 copyright protection537ＰＥＮＡＮＡBDbgsj8nj5 維尼

f=cos(u+a),g′=eu8964 copyright protection537ＰＥＮＡＮＡxeEOv3SWZE 維尼

↓ steps↓ steps8964 copyright protection537ＰＥＮＡＮＡ1L99CtHelg 維尼

f′=−sin(u+a),g=eu:8964 copyright protection537ＰＥＮＡＮＡsJj6PQObm1 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection537ＰＥＮＡＮＡTZIgY5Cw3L 維尼

Apply linearity:8964 copyright protection537ＰＥＮＡＮＡATScuCDJlP 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection537ＰＥＮＡＮＡtqeIvqoaVx 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection537ＰＥＮＡＮＡvxjYn7MKfs 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection537ＰＥＮＡＮＡHbxcoLjr4G 維尼

Plug in solved integrals:8964 copyright protection537ＰＥＮＡＮＡX2V9lV9G3o 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection537ＰＥＮＡＮＡNFWbZLASHp 維尼

Undo substitution u=√x:8964 copyright protection537ＰＥＮＡＮＡhnnHFIaG22 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection537ＰＥＮＡＮＡ9CjvsVDnqN 維尼

The problem is solved:8964 copyright protection537ＰＥＮＡＮＡNZGp9XorGv 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection537ＰＥＮＡＮＡFQtvDsIpvH 維尼

Rewrite/simplify:8964 copyright protection537ＰＥＮＡＮＡGLaXOYEbz3 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection537ＰＥＮＡＮＡfjpHmHkbMp 維尼

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