 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/17/a/iii/solution

x = sin u

with

- π /2 \leq u \leq π /2

. The cosine is nonnegative there, so

1 - x^{2} = cos u

and

d x = cos u d u

\int 1 - x^{2} d x = \int cos^{2} u d u = \frac{u}{2} + \frac{1}{4} sin 2 u + C .

Since

u = arcsin x

and

sin 2 u = 2 x 1 - x^{2}

\int 1 - x^{2} d x = \frac{1}{2} (arcsin x + x 1 - x^{2}) + C .

(- 1, 1)

and has continuous endpoint values at

\pm 1

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