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恐懼感8964 copyright protection418ＰＥＮＡＮＡFJqJHrsUkj 維尼

422Please respect copyright.ＰＥＮＡＮＡiKkhIU8sl4
8964 copyright protection418ＰＥＮＡＮＡcIL6wLbiq3 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection418ＰＥＮＡＮＡtWyRPPU1PB 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection418ＰＥＮＡＮＡuS9aVpZOYu 維尼

=2∫eusin(u+a)du… or choose an alternative:422Please respect copyright.ＰＥＮＡＮＡwM8R9gxTGB
Substitute e√x8964 copyright protection418ＰＥＮＡＮＡg0a3Ko7DkQ 維尼

Now solving:8964 copyright protection418ＰＥＮＡＮＡaGjV0n1wjw 維尼

∫eusin(u+a)du8964 copyright protection418ＰＥＮＡＮＡ6IAHKXG4iu 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection418ＰＥＮＡＮＡt2Bx97jdAi 維尼

First time:8964 copyright protection418ＰＥＮＡＮＡiDVXQyYz8k 維尼

f=sin(u+a),g′=eu8964 copyright protection418ＰＥＮＡＮＡgimJFmv0B2 維尼

↓ steps↓ steps8964 copyright protection418ＰＥＮＡＮＡvZS7kDRscf 維尼

f′=cos(u+a),g=eu:8964 copyright protection418ＰＥＮＡＮＡSbJkekNDhC 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection418ＰＥＮＡＮＡv4nyYj0OZI 維尼

Second time:8964 copyright protection418ＰＥＮＡＮＡ2g9LO8MhEk 維尼

f=cos(u+a),g′=eu8964 copyright protection418ＰＥＮＡＮＡYe1cjnRR3p 維尼

↓ steps↓ steps8964 copyright protection418ＰＥＮＡＮＡafyWFDbaHg 維尼

f′=−sin(u+a),g=eu:8964 copyright protection418ＰＥＮＡＮＡcQebmIbePj 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection418ＰＥＮＡＮＡzabqMSh4mF 維尼

Apply linearity:8964 copyright protection418ＰＥＮＡＮＡlIBCmcKL8a 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection418ＰＥＮＡＮＡ7hqpAAJslm 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection418ＰＥＮＡＮＡVTXAYN0dba 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection418ＰＥＮＡＮＡXVf8CXl1Hj 維尼

Plug in solved integrals:8964 copyright protection418ＰＥＮＡＮＡYsGmIrV669 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection418ＰＥＮＡＮＡIIXufTKBJn 維尼

Undo substitution u=√x:8964 copyright protection418ＰＥＮＡＮＡuQWBiKQd7C 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection418ＰＥＮＡＮＡgyB5u8ewl7 維尼

The problem is solved:8964 copyright protection418ＰＥＮＡＮＡ7bea2xELY8 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection418ＰＥＮＡＮＡCra1bmx0MT 維尼

Rewrite/simplify:8964 copyright protection418ＰＥＮＡＮＡzKoQ41U3d2 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection418ＰＥＮＡＮＡAr1OwUdr5P 維尼

3.12.34.113