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Math 180 Packet #20 Substitution
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The idea for the substitution method is to make a substitution so that the integral is simpler.
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Ex 1. Evaluate: π₯ 3 +π₯ 5 3 π₯ 2 +1 ππ₯
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The Substitution Rule If π’=π(π₯) is a differentiable function whose range is an interval πΌ, and π is continuous on πΌ, then π π π₯ π β² π₯ ππ₯= π π’ ππ’
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Proof: Suppose πΉ β² =π and π’=π(π₯)
Proof: Suppose πΉ β² =π and π’=π(π₯). Then, π π π₯ π β² π₯ ππ₯ = π ππ₯ πΉ π π₯ ππ₯ =πΉ π π₯ +πΆ =πΉ π’ +πΆ = πΉ β² π’ ππ’ = π π’ ππ’
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Proof: Suppose πΉ β² =π and π’=π(π₯)
Proof: Suppose πΉ β² =π and π’=π(π₯). Then, π π π₯ π β² π₯ ππ₯ = π ππ₯ πΉ π π₯ ππ₯ =πΉ π π₯ +πΆ =πΉ π’ +πΆ = πΉ β² π’ ππ’ = π π’ ππ’
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Proof: Suppose πΉ β² =π and π’=π(π₯)
Proof: Suppose πΉ β² =π and π’=π(π₯). Then, π π π₯ π β² π₯ ππ₯ = π ππ₯ πΉ π π₯ ππ₯ =πΉ π π₯ +πΆ =πΉ π’ +πΆ = πΉ β² π’ ππ’ = π π’ ππ’
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Proof: Suppose πΉ β² =π and π’=π(π₯)
Proof: Suppose πΉ β² =π and π’=π(π₯). Then, π π π₯ π β² π₯ ππ₯ = π ππ₯ πΉ π π₯ ππ₯ =πΉ π π₯ +πΆ =πΉ π’ +πΆ = πΉ β² π’ ππ’ = π π’ ππ’
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Proof: Suppose πΉ β² =π and π’=π(π₯)
Proof: Suppose πΉ β² =π and π’=π(π₯). Then, π π π₯ π β² π₯ ππ₯ = π ππ₯ πΉ π π₯ ππ₯ =πΉ π π₯ +πΆ =πΉ π’ +πΆ = πΉ β² π’ ππ’ = π π’ ππ’
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Proof: Suppose πΉ β² =π and π’=π(π₯)
Proof: Suppose πΉ β² =π and π’=π(π₯). Then, π π π₯ π β² π₯ ππ₯ = π ππ₯ πΉ π π₯ ππ₯ =πΉ π π₯ +πΆ =πΉ π’ +πΆ = πΉ β² π’ ππ’ = π π’ ππ’
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Notes: To use the substitution method, what youβre looking for is a function and its derivative. The function for π’ will usually be inside another function. You want the substitution to make the new integral easier to integrate.
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Ex 2. Evaluate: sec 2 (5π₯+1) β
5 ππ₯ Ex 3. Evaluate: 2π₯+1 ππ₯ Ex 4
Ex 2. Evaluate: sec 2 (5π₯+1) β
5 ππ₯ Ex 3. Evaluate: 2π₯+1 ππ₯ Ex 4. Evaluate: π₯ 2 π π₯ 3 ππ₯
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Sometimes you need to do some algebra before you can do a substitution
Sometimes you need to do some algebra before you can do a substitution. Ex 5. Evaluate: ππ₯ π π₯ + π βπ₯ Ex 6. Evaluate: sec π₯ ππ₯
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Ex 7. Evaluate: π₯ 2π₯+1 ππ₯
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