# Integration of function (inverse)

Does anyone know how do I start on part (b)?

Thanks

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Hint: let $u = \sqrt{1-x}$
or $x = \sin^2t$
For integrands that involve a square root (and the argument of the square root isn't a sum or difference of squares, which usually suggests a trigonometric substitution), it often works simply to substitute using the radical quantity as the new variable, e.g., in this case $u = \sqrt{1 - x}$, in part because this often rationalizes the expression. Rearranging gives $x = 1 - u^2$ and so $dx = -2u du$.
Alternately, you can substitute $x = \sin^2 t$.