121616949-math.171 - idea is simple guess at the most...

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8.1 Substitution 157 This looks messy, but we do now have something that looks like the result of the chain rule: the function 1 - x 2 has been substituted into - (1 / 2)(1 - x ) x , and the derivative of 1 - x 2 , - 2 x , multiplied on the outside. If we can find a function F ( x ) whose derivative is - (1 / 2)(1 - x ) x we’ll be done, since then d dx F (1 - x 2 ) = - 2 xF (1 - x 2 ) = ( - 2 x ) - 1 2 (1 - (1 - x 2 )) p 1 - x 2 = x 3 p 1 - x 2 But this isn’t hard: Z - 1 2 (1 - x ) x dx = Z - 1 2 ( x 1 / 2 - x 3 / 2 ) dx ( 8 . 1 ) = - 1 2 2 3 x 3 / 2 - 2 5 x 5 / 2 + C = 1 5 x - 1 3 x 3 / 2 + C. So finally we have Z x 3 p 1 - x 2 dx = 1 5 (1 - x 2 ) - 1 3 (1 - x 2 ) 3 / 2 + C. So we succeeded, but it required a clever first step, rewriting the original function so that it looked like the result of using the chain rule. Fortunately, there is a technique that makes such problems simpler, without requiring cleverness to rewrite a function in just the right way. It does sometimes not work, or may require more than one attempt, but the
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Unformatted text preview: idea is simple: guess at the most likely candidate for the “inside function”, then do some algebra to see what this requires the rest of the function to look like. One frequently good guess is any complicated expression inside a square root, so we start by trying u = 1-x 2 . We use a new variable, u , for convenience in the manipulations that follow. Now we know that the chain rule will multiply by the derivative of this inner function: du dx =-2 x, so we need to rewrite the original function to include this: Z x 3 p 1-x 2 = Z x 3 √ u-2 x-2 x dx = Z x 2-2 √ u du dx dx....
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