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恐懼感8964 copyright protection444ＰＥＮＡＮＡrVbB4vAg37 維尼

448Please respect copyright.ＰＥＮＡＮＡ7dEAPMpBAk
8964 copyright protection444ＰＥＮＡＮＡdhIK8fhYQr 維尼

∫sin(√x+a)e√x√xdx8964 copyright protection444ＰＥＮＡＮＡvsatJQOnsN 維尼

Substitute u=√x ⟶ dudx=12√x (steps) ⟶ dx=2√xdu:8964 copyright protection444ＰＥＮＡＮＡNrZkHX4Ilv 維尼

=2∫eusin(u+a)du… or choose an alternative:448Please respect copyright.ＰＥＮＡＮＡ12IeHqSZqu
Substitute e√x8964 copyright protection444ＰＥＮＡＮＡs1CDhbgyVY 維尼

Now solving:8964 copyright protection444ＰＥＮＡＮＡYsydlyNCaU 維尼

∫eusin(u+a)du8964 copyright protection444ＰＥＮＡＮＡePijQIRycu 維尼

We will integrate by parts twice in a row: ∫fg′=fg−∫f′g.8964 copyright protection444ＰＥＮＡＮＡRYVLlr0vut 維尼

First time:8964 copyright protection444ＰＥＮＡＮＡPMYg38L2dm 維尼

f=sin(u+a),g′=eu8964 copyright protection444ＰＥＮＡＮＡy4wXF4syF8 維尼

↓ steps↓ steps8964 copyright protection444ＰＥＮＡＮＡvqNjCRgppQ 維尼

f′=cos(u+a),g=eu:8964 copyright protection444ＰＥＮＡＮＡ0AiU71FvyK 維尼

=eusin(u+a)−∫eucos(u+a)du8964 copyright protection444ＰＥＮＡＮＡl9kp1s4S5r 維尼

Second time:8964 copyright protection444ＰＥＮＡＮＡABwoE5Lj0S 維尼

f=cos(u+a),g′=eu8964 copyright protection444ＰＥＮＡＮＡfXiy8CQNX9 維尼

↓ steps↓ steps8964 copyright protection444ＰＥＮＡＮＡN825LuXGIR 維尼

f′=−sin(u+a),g=eu:8964 copyright protection444ＰＥＮＡＮＡhJ7OQbfF4B 維尼

=eusin(u+a)−(eucos(u+a)−∫−eusin(u+a)du)8964 copyright protection444ＰＥＮＡＮＡoRthsqeDCU 維尼

Apply linearity:8964 copyright protection444ＰＥＮＡＮＡeNNXyP16kS 維尼

=eusin(u+a)−(eucos(u+a)+∫eusin(u+a)du)8964 copyright protection444ＰＥＮＡＮＡaUCLha9blw 維尼

The integral ∫eusin(u+a)du appears again on the right side of the equation, we can solve for it:8964 copyright protection444ＰＥＮＡＮＡQChOjyfnem 維尼

=eusin(u+a)−eucos(u+a)28964 copyright protection444ＰＥＮＡＮＡ6E0CNCukRX 維尼

Plug in solved integrals:8964 copyright protection444ＰＥＮＡＮＡat9i9XcMNw 維尼

2∫eusin(u+a)du=eusin(u+a)−eucos(u+a)8964 copyright protection444ＰＥＮＡＮＡqrW1r3mTma 維尼

Undo substitution u=√x:8964 copyright protection444ＰＥＮＡＮＡLrcJAtpuUT 維尼

=sin(√x+a)e√x−cos(√x+a)e√x8964 copyright protection444ＰＥＮＡＮＡ34ReWvYUzM 維尼

The problem is solved:8964 copyright protection444ＰＥＮＡＮＡuD99mUCcw1 維尼

∫sin(√x+a)e√x√xdx=sin(√x+a)e√x−cos(√x+a)e√x+C8964 copyright protection444ＰＥＮＡＮＡ6tBMqdbnDt 維尼

Rewrite/simplify:8964 copyright protection444ＰＥＮＡＮＡdfcDhLfs3k 維尼

=(sin(√x+a)−cos(√x+a))e√x+C8964 copyright protection444ＰＥＮＡＮＡaLS1uykVww 維尼

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