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103

Part of 2006 Today's Calculation Of Integral

Problems(1)

Today's calculation of integral 103

Source: Shizuoka University entrance exam 1988

6/5/2006
For 0<a<1,0<a<1, let f(x)=a−x1−ax (−1<x<1).f(x)=\frac{a-x}{1-ax}\ (-1<x<1). Evaluate ∫0a1−{f(x)}61−x2 dx.\int_0^a \frac{1-\{f(x)\}^6}{1-x^2}\ dx.
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