How do I integrate [math]v^3(a^2-v^2)^{\frac{1}{2}}[/math] by substitution?
To integrate ( v^3(a^2-v^2)^{\frac{1}{2}} ) by substitution, you can follow these steps:
Step-by-Step Integration
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Choose a Substitution: Let ( u = a^2 - v^2 ). This means ( du = -2v \, dv ), or ( v \, dv = -\frac{1}{2} du ).
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Express ( v^3 ) in Terms of ( u ): Since ( u = a^2 - v^2 ), we have ( v^2 = a^2 - u ). Thus, ( v^3 = v \cdot v^2 = v(a^2 - u) ).
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Transform the Integral: Substitute ( u ) and ( du ) into the integral. The integral becomes: [ \int v^3(a^2-v^2)^{\frac{1}{2}} \, dv = \int v(a^2 - u)^{\frac{1}{2}} \cdot v \cdot -\frac{1}{2} \, du ] Simplify this to: [ -\frac{1}{2} \int (a^2 - u) \cdot u^{\frac{1}{2}} \, du ]
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Expand and Integrate: Expand the integral and integrate term by term: [ -\frac{1}{2} \int (a^2u^{\frac{1}{2}} - u^{\frac{3}{2}}) \, du ] Integrating each term gives: [ -\frac{1}{2} \left( a^2 \cdot \frac{2}{3}u^{\frac{3}{2}} - \frac{2}{5}u^{\frac{5}{2}} \right) + C ] Simplify to: [ -\frac{1}{3}a^2u^{\frac{3}{2}} + \frac{1}{5}u^{\frac{5}{2}} + C ]
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Substitute Back: Replace ( u ) with ( a^2 - v^2 ): [ -\frac{1}{3}a^2(a^2 - v^2)^{\frac{3}{2}} + \frac{1}{5}(a^2 - v^2)^{\frac{5}{2}} + C ]
This is the integrated form of the given function using substitution.